Help me with this math problem please

Two functions that satisfy the given condition and are linearly independent on (-∞,∞) are f1(x) = e^(x) and f2(x) = e^(x)

Let's start by recalling the definition of the Wronskian for two functions f1(x) and f2(x):

W(f1,f2)(x) = f1(x)f2'(x) - f1'(x)f2(x)

Given the Wronskian W(f1,f2)(x) = e^(2x), we can try to find functions f1(x) and f2(x) that satisfy this condition. One possible solution is:

f1(x) = e^(x)

f2(x) = e^(x + C)

where C is a constant. We can now calculate the Wronskian of f1(x) and f2(x):

W(f1,f2)(x) = f1(x)f2'(x) - f1'(x)f2(x)

= e^(x) [e^(x + C)]' - [e^(x)]' e^(x + C)

= e^(x) e^(x + C) - e^(x) e^(x + C)

= 0

This means that f1(x) and f2(x) are linearly dependent. However, we can modify our choice of f2(x) by setting C= -x, which gives:

f2(x) = e^(2x - x) = e^(x)

Now we can calculate the Wronskian again:

W(f1,f2)(x) = f1(x)f2'(x) - f1'(x)f2(x)

= e^(x) [e^(x)]' - [e^(x)]' e^(x)

= e^(2x)

Since the Wronskian is non-zero, this means that f1(x) and f2(x) are linearly independent on (-∞,∞).

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The given question is incomplete, the complete question is:

Determine two functions, defined on the interval (−∞,∞), whose Wronskian is given by W(f 1 ,f 2)=e ^2x. Are the functions that you found linearly independent on (−∞,∞)

81 must be the bare minimum acceptable sample size for a real-world computation.

The mean of a group of two or more integers is the straightforward mathematical average. The geometric mean approach, which uses the average of a set of products, and the arithmetic mean technique, which uses the sum of the series' values, are only two of the techniques available to calculate the mean for a given set of data.

Given: For the population of healthy people, the mean IQ score is μ =100 and the standard deviation is б =14.

Significance level : 1-0.99=.01

Critical value : zₐ/2=2.576

Margin of error : E=4

Standard deviation : б =14

The following equation determines the sample size:

n=( zₐ/2×б /E)²

n=(2.576×14/4)²

n=81.288≈81

Hence, the minimum reasonable sample size for a real world calculation must be 81.

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The probability that the student selected to work on a problem on the board is C. 0. 20.

The total number of students in Junior grade can be found to be 11 by the graph. Out of this 11, the number of female students is:

= 11 - 6

= 5 students

The total number of students in the survey is:

= 7 sophomore + 11 junior + 7 senior

= 25 students

The probability that a female junior student is chosen is:

= 5 / 25 x 100 %

= 1 / 5 x 100 %

= 0. 20

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Let's denote the radius of the cone as 'r' and the height as 'h'.

The formula for the volume of a cone is:

V = (1/3) * π * r^2 * h

And the formula for the circumference of the base of a cone is:

C = 2 * π * r

We know that the volume of the cone is 33π/3, so:

(1/3) * π * r^2 * h = 33π/3

Simplifying this equation, we get:

π * r^2 * h = 99π

r^2 * h = 99

We also know that the circumference of the base of the cone is 6π, so:

2 * π * r = 6π

Simplifying this equation, we get:

r = 3

Now we can substitute this value of 'r' into the equation we obtained earlier:

r^2 * h = 99

3^2 * h = 99

h = 11

Therefore, the height of the cone is 11 units.

A sample of 312 bacteria would be needed to estimate the mean lifespan of the species of bacteria within a margin of error of 0.75 hours with 95% confidence.

To determine the sample size required to estimate the mean lifespan of a certain species of bacteria within a margin of error of 0.75 hours with a 95% level of confidence, we can use the following formula:

n = [Z*(σ/ME)]^2

where:

n = sample size

Z = the z-score corresponding to the desired level of confidence, which is 1.96 for a 95% level of confidence

σ = the population standard deviation

ME = the desired margin of error

From the given information, we have:

sample mean = Xbar = hours

sample size = n = 35

sample standard deviation = s = hours

desired margin of error = ME = 0.75 hours

level of confidence = 95%, which corresponds to a z-score of 1.96

We do not know the population standard deviation, but we can use the sample standard deviation as an estimate since the sample size is greater than 30. Thus, we have:

σ ≈ s = hours

Substituting the values into the formula, we get:

n = [1.96*(6.81/0.75)]^2 ≈ 311.84

Rounding up to the nearest whole number, we get a sample size of 312 bacteria.

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Keith found that the product of 31.5 and 5.3 is 160.15

Keith's strategy for finding the product is breaking the numbers into parts and then adding the results.

The product of 21.5 and 5.3 is

(20 + 1 + 0.5) × (5 + 0.3) = (20 × 5) + (20 × 0.3) + (1 × 5) + (1 × 0.3) + (0.5 × 5) + (0.5 × 0.3) = 100 + 6 + 5 + 0.3 + 2.5 + 0.15 = 114.95

When multiplying 31.5 by 5.3, Keith used a similar approach, breaking the numbers into parts and then adding the results. Specifically, he expanded the terms as (31 + 0.5)(5 + 0.3), added the partial products, and then rounded the result to two decimal places.

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The 95% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is -36.54 to -31.26.

To calculate the 95% confidence interval for the true difference between the testing averages of the two instructional methods, we can use the following formula

Confidence interval = (X1 - X2) ± Z*(σ1^2/n1 + σ2^2/n2)^0.5

Where

X1 is the testing average for Method 1

X2 is the testing average for Method 2

Z is the critical value for a 95% confidence interval, which is 1.96

σ1 is the known standard deviation for Method 1

σ2 is the known standard deviation for Method 2

n1 is the sample size for Method 1

n2 is the sample size for Method 2

Substituting the values given in the problem, we get

Confidence interval = (55.9 - 89.8) ± 1.96*(8.92^2/179 + 16^2/176)^0.5

= -33.9 ± 2.64

Therefore, the 95% confidence interval using Method 1 and students using Method 2 is -36.54 to -31.26.

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The given question is incomplete, the complete question is:

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 179 students using Method 1 produces a testing average of 55.9. A sample of 176 students using Method 2 produces a testing average of 89.8. Assume the standard deviation is known to be 8.92 for Method 1 and 16 for Method 2. Determine the 95% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2

Answer: 3.2

Step-by-step explanation: put the 9.4 over the 6.2 (not as a fraction) subtract and then you get 0.2 when you take the .4 - .2 and 9 - 6 is 3 then you would put the 0.2 with the 3 to get 3.2

The first term of the geometric progression is 1.

What is geometric progression?

Each term in the sequence preceding it is multiplied by a fixed number called a common ratio to produce the next phrase in a series known as a geometric progression (GP). A geometric number sequence that follows a pattern is another name for this progression.

We are given a geometric progression as b₁, b₂, 4, -8, ....

From this, we get the common ratio as [tex]\frac{-8}{4}[/tex], which is -2.

Now, we know that

a₃ = 4 and r = -2

So,

⇒a₃ = a₁ [tex]r^{n-1}[/tex]

⇒4 = a₁ * [tex]-2^ {3-1}[/tex]

⇒4 = 4a₁

⇒a₁ = 1

Hence, the first term of the geometric progression is 1.

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Since, there are multiple questions, so the question answered above is:

Question: Find the first term of the geometric progression:

b₁, b₂, 4, -8, ...

a) 1; b) -1; c) 28; d) [tex]\frac{1}{2}[/tex]

3/5 divided by 2/3 is equal to 9/10. We can also simplify this fraction by dividing both the numerator and the denominator by their greatest common factor, which is 1 in this case.

To divide fractions, we need to invert the second fraction and then multiply the two fractions together. This can be written as:

(a/b) ÷ (c/d) = (a/b) × (d/c)

In this case, a = 3, b = 5, c = 2, and d = 3. Substituting these values into the above equation, we get:

(3/5) ÷ (2/3) = (3/5) × (3/2)

To multiply fractions, we simply multiply their numerators together and their denominators together. So, we can rewrite the above equation as:

(3/5) × (3/2) = (3 × 3) / (5 × 2) = 9/10

Therefore, 3/5 divided by 2/3 is equal to 9/10. We can also simplify this fraction by dividing both the numerator and the denominator by their greatest common factor, which is 1 in this case. Therefore, the final answer is 9/10.

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the greatest common factor of the given expressions is -5xy.

To find the greatest common factor of the given expressions, we need to factor them first.

[tex]-15x^2y - 10xy^3 + 5xy^4 = -5xy(3x^2 + 2y^2 - y^3)\\5xy = 5xy(1)\\5x^2y^4 = 5xy^4(x^2)\\5xy^2 = 5xy^2(1)[/tex]

xy = xy(1)

Now, we can find the common factors by taking the minimum exponent of each variable in the factors:

The common factors are:5xy

Therefore, the greatest common factor of the given expressions is -5xy.

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The value of x, considering the similar triangles, is given as follows:

x = 26.

Similar triangles are triangles that share these two features listed as follows:

The two triangles in this problem are similar, due to the bisection, hence the equivalent side lengths are given as follows:

Then the proportional relationship is given as follows:

x/13 = 10/5.

Meaning that the value of x is given as follows:

x/13 = 2

x = 26.

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The equations of the functions are x = -10, y = x + 5 and y = x² + 6x + 10, and the identities are shown below

Constant function

This function remains constant regardless of the input and/or output.

The graph shows a vertical line located at x = -10.

The function represented by this graph is constant, with a domain of x = -10 and a range of y [0, 12.25]

Linear function

The function varies continuously in relation to both x and y.

We can observe two points on the graph, which are (-4, 1) and (-2, 3).

From the point, we can see that y is more than x by 5

This means that the function is y = x + 5

The following identities of the functions are Domain: [-4, -2] and Range: [1, 3]

Quadratic function

This function can be represented as

y = a(x - h)² + k

From the graph, we have

(h, k) = (-3, 1) and (x, y) = (-2, 2)

By substitution, we have

y = a(x + 3)² + 1

By substitution, we have

2 = a(-2 + 3)² + 1

Solving for a, we have

a = 1

So, we have

y = (x + 3)² + 1

y = x² + 6x + 10

This means that the function is y = x² + 6x + 10 with the following identities

Domain: [-4, -2]

Range: [1. 2]

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Answer: #4 = Non-Linear & #3 = Linear

Step-by-step explanation: Non-Linear lines create curves and not straight lines. As the term non-"line"ar defines not as a line, linear represent straight lines. Hence, #4 is non-linear and #3 is linear in terms of graphing.

Answer:

If 2inches= 20 miles then 1inches =10miles

5.6 inches=5.6*10

=56miles

Pls ad me brainliest.

Answer:

56 miles

Step-by-step explanation:

Use a proportion.

2 inches is to 20 miles as 5.6 inches is to x miles.

2:20 = 5.6:x

2/20 = 5.6/x

1/10 = 5.6/x

1 × x = 10 × 5.6

Cross multiply.

x = 56

Answer: 56 miles

Answer:

15

Step-by-step explanation:

Place the number in order from least to greatest. 13,14,14,16,16,19

then take one away from each side till there is one left. 14,14,16,16 14,16.

in this case there is two from this point on you find the average of the two number you have left so in this case, 14+16=30 30/2=15 The median is 15

Answer:

187

Step-by-step explanation:

If 70th after he see 84 and legs 33 so the answer is 70+84+33=187

1) The line number and the plot of the numbers are shown in the graph that is attached.

2) The opposite of an integer is offset by the same amount from zero.

The three numerals on the number line and their corresponding opposites are shown in the pictures that are attached.

We must determine the number of places on the scale starting at zero to determine a number's opposite. The same amount of places is then counted starting from zero in the opposite direction.

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Complete question:

Draw a number line, and create a scale for the number line to plot the points −2, 4, and 6.

a. Graph each point and its opposite on the number line.

b. Explain how you found the opposite of each point.

Answer:

It cannot be in the sequence because counting back in 0.4 will give you even numbers and not odd hence 1.5 cannnot be in the sequence.

PLEASE MARK IT THE BRAINLIEST!!!

Answer:

Find the nth term of 23,17,11,5:

23,17,11,5,-1,-7,-13...

The repeating decimal produced by the fraction 3/9 is 0.33333..., with the digit 3 repeating infinitely.

A fraction is a way of expressing a part of a whole or a ratio between two quantities. It is represented as one quantity divided by another quantity, with a horizontal line called a fraction bar between them.

The fraction 3/9 can be simplified to 1/3 by dividing both the numerator and denominator by their greatest common factor, which is 3.

When we divide 1 by 3, we get a quotient of 0.3, and a remainder of 1. To continue the long division, we add a decimal point and a zero, and bring down the next digit, which is also a zero. We then divide 10 by 3, which gives us a quotient of 3, and a remainder of 1. We repeat the process, adding a decimal point and another zero, and bringing down another zero. We continue this process infinitely, getting a sequence of 3s that repeat without end.

Therefore, the repeating decimal produced by the fraction 3/9 is 0.33333..., with the digit 3 repeating infinitely.

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Answer:1 hour 15 Minutes

Step-by-step explanation:The glider begins its flight mile above the ground.

Distance above the ground after 45 minutes =

Change in height of the glider

Next, we determine how long the entire descent last.

Expressing the distance moved as a ratio of time taken

Therefore: Total Time taken =45+30=75 Minutes

=1 hour 15 Minutes

the total number of coins in the museum's collection is:-144 + 90 + 60 + 30 + 10 + 5 = 339 coins.

What is histogram ?

A histogram is a graphical representation of the distribution of a dataset. It is commonly used in statistics to represent the frequency distribution of a set of continuous or discrete data. In a histogram, the data is divided into intervals, or bins, and the frequency of the data falling into each bin is represented by the height of a bar.

The x-axis of a histogram represents the range of values in the dataset, while the y-axis represents the frequency or count of data points falling within each bin. The bars of a histogram are usually drawn touching each other, as the data is continuous and there are no gaps between the bins.

To determine the total number of Roman coins in the museum's collection, we need to know the area under the histogram.

Since we know that 144 coins each weigh between 8 g and 17 g, we can calculate the total weight of those coins:

144 coins x ((17 g - 8 g)/2) = 144 coins x 4.5 g = 648 g

This means that the area of the rectangle representing those coins in the histogram is:

144 coins x 9 g = 1296 g

To find the total number of coins, we need to calculate the area of the remaining rectangles in the histogram. Since the width of each rectangle is 9 g, we can calculate the height of each rectangle by dividing its area by 9 g.

The second rectangle has an area of:

90 coins x 9 g = 810 g

So its height is:

810 g / 9 g = 90 coins

The third rectangle has an area of:

60 coins x 9 g = 540 g

So its height is:

540 g / 9 g = 60 coins

The fourth rectangle has an area of:

30 coins x 9 g = 270 g

So its height is:

270 g / 9 g = 30 coins

The fifth rectangle has an area of:

10 coins x 9 g = 90 g

So its height is:

90 g / 9 g = 10 coins

Finally, the sixth rectangle has an area of:

5 coins x 9 g = 45 g

So its height is:

45 g / 9 g = 5 coins

Therefore, the total number of coins in the museum's collection is:

144 + 90 + 60 + 30 + 10 + 5 = 339 coins.

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Answer:

[tex]\frac{2xx^{\frac{1}{4} } }{y^{\frac{11}{3} } }[/tex]

Step-by-step explanation:

Answer:

2100 km

Step-by-step explanation:

If the entire trip was 2400 km, then we can note that x km was flown by plane, and 2400 - x km by bus

The whole trip took 8 hours and we know that t = s/v:

[tex]t(by \: plane) = \frac{x}{700} [/tex]

[tex]t(by \: bus) = \frac{2400 - x}{60} [/tex]

Now we can form an equation:

[tex] \frac{x}{700} + \frac{2400 - x}{60} = 8 [/tex]

[tex] \frac{60x + 1680000 - 700x}{42000} = 8[/tex]

[tex] \frac{ - 640x + 1680000}{42000} = 8 [/tex]

Use the property of the proportion:

-640x + 1680000 = 336000

-640x = 336000 - 1680000

-640 x = -1344000 / : (-640)

x = 2100

Since we've noted that x is a path flew by the plane, we have the answer already

Answer:

Step-by-step explanation:

16.15 is the value of ZY in triangle .

What is known as a triangle?

The three vertices of a triangle make it a three-sided polygon. The angles of the triangle are formed by the connection of the three sides end to end at a single point.

                        The triangle's three angles add up to 180 degrees. There are three straight sides to this two-dimensional shape. An example of a 3-sided polygon is a triangle. Three triangle angles added together equal 180 degrees.

In ΔUST and ΔZXY

     UT/ZY =  UA/ZB

     8.5/ZY =  6/11.4

     6 *  ZY =  11.4 * 8.5

            ZY = 96.9/6

                  = 16.15

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The probability of being dealt 4 hearts from a standard 52 card deck if only 4 cards are to be dealt is 0.0026.

There are 52 cards in a standard deck, 13 of which are hearts.

We want to find the probability of getting dealt four hearts out of a four-card hand.

Since order doesn't matter, we will use combinations to calculate this probability.

P(A) = number of favorable outcomes / total number of possible outcomes.

There are 13 choose 4 ways to choose four hearts from the deck, and there are 52 choose 4 ways to choose any four cards from the deck.

So the probability of being dealt four hearts from a standard 52 card deck if only 4 cards are to be dealt is:

P(A) = (13 choose 4) / (52 choose 4)P(A)

       = (715) / (270725)P(A)

       = 0.002641056.

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4. CE is a median of triangle ADF,  BF is the midpoint of AC.

5. CE = 3√3 cm and AG = 18√3 cm

6. AM is a median of right triangle ABC.

In mathematics, a triangle is a geometric shape that consists of three line segments that intersect at three endpoints. These endpoints are called vertices, and the line segments are called sides.

Triangles are one of the most basic shapes in geometry and are used in many areas of mathematics, science, engineering, and everyday life. They can be classified based on the length of their sides and the measure of their angles.

4. CE is a median of triangle ADF.

BF is the midpoint of AC.

Since C is the midpoint of segment AD and E is the midpoint of segment FD, CE is a median of triangle ADF. This is because CE passes through D and divides the opposite side A F into two equal halves.

Similarly, since B is the midpoint of segment AC, BF is a midpoint of AC. This is because BF passes through A and divides the opposite side AC into two equal halves.

5. To find CE and AG, we can use the fact that BF = 24 cm. Let's start by finding CE.

Since BF is the midpoint of AC, we have:

AC = 2 BF = 2(24 cm) = 48 cm.

Since C is the midpoint of AD, we have:

AD = 2 CD = 2 CE (by definition of midpoint)

Therefore, CE = AD/2

To find AD, we can use the fact that E is the midpoint of FD, so:

FD = 2 FE = 2 FG (by definition of midpoint)

Since F is the midpoint of GE, we have:

GE = 2 GF (by definition of midpoint)

Therefore, AG = AE + GE = AE + 2GF

Now, since AM is a median of triangle ABC, we have:

AM² = (AB² + BC²)/2 - (AC²/4)

Since triangle ABC is a right-angled triangle with angle B = 90 degrees, we have:

AB² + BC² = AC²

Therefore, we can simplify the equation for AM² as follows:

AM² = AC²/2 - AC²/4

AM² = AC²/4

Substituting the value we found for AC, we get:

AM² = 48²/4 = 576

Therefore, AM = 24 cm.

Now, we can use the Pythagorean theorem to find AE:

AE² + EM² = AM²

AE² + (AC/2)² = AM²

AE² + 24² = 576

AE² = 576 - 576/4 = 432

AE = √432 = 12√3 cm

Finally, we can find AG as:

AG = AE + 2GF = AE + 2(AD/4) (since G is the midpoint of EF)

AG = 12√3 + AD/2

But we know that AD = 2CE, so:

AG = 12√3 + CE

Therefore, to find AG, we just need to add the value of CE that we found earlier:

AG = 12√3 + CE = 12√3 + (AD/2) = 12√3 + (CE/2) = 12√3 + 6√3 = 18√3 cm.

Therefore, CE = AD/2 = (AG - 12√3)/2 = (18√3 - 12√3)/2 = 3√3 cm.

So, CE = 3√3 cm and AG = 18√3 cm.

6. Statement: AM is a median of right triangle ABC.

A median of a triangle is a line segment that joins a vertex to the midpoint of the opposite side. In right triangle ABC, the median AM joins the right angle vertex A to the midpoint of the hypotenuse BC.

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4). Section AC's midpoint is B, and section BF's midpoint is BF. Because BF cuts through A and splits the opposing side AC into two equal halves, this is the case.

5). CE = AD/2 = (AG - 12√3)/2 = (18√3 - 12√3)/2 = 3√3 cm.

So, CE = 3√3 cm and AG = 18√3 cm.

6). The middle of the hypotenuse BC is connected to the right angle vertex A by the median AM.

One of the most fundamental geometric shapes, triangles are used frequently in mathematics, science, engineering, and daily living. They can be categorized based on the dimensions of their edges and sides.

4). ADF's triangle's middle is CE.

BF sits in the middle of AC.

Triangle ADF's median is CE because C is the midway of segment AD and E is the midpoint of segment FD. This is the case because CE passes through D and divides the opposite side A F into two equal halves.

BF functions as an AC midpoint in a manner similar to how B acts as the segment's halfway point. This is the case because BF cuts through A and divides the opposite side AC into two equal halves.

5). We can use the knowledge that BF = 24 centimeters to determine CE and AG. Let's begin by locating CE.

Because BF is AC's middle, we have:

AC = 2 BF = 2(24 cm) = 48 cm.

Since C is the midpoint of AD, we have:

AD = 2 CD = 2 CE (by definition of midpoint)

Therefore, CE = AD/2

To find AD, we can use the fact that E is the midpoint of FD, so:

FD = 2 FE = 2 FG (by definition of midpoint)

Since F is the midpoint of GE, we have:

GE = 2 GF (by definition of midpoint)

Therefore, AG = AE + GE = AE + 2GF

Now that triangle ABC's middle is AM, we have:

AM² = (AB² + BC²)/2 - (AC²/4)

Triangle ABC has a right angle of 90 degrees, so we have:

AB² + BC² = AC²

As a result, we can simplify the AM² equation as follows:

AM² = AC²/2 - AC²/4

AM² = AC²/4

When we substitute the AC number we discovered, we obtain:

AM² = 48²/4 = 576

Therefore, AM = 24 cm.

We can now determine AE using the Pythagorean theorem:

AE² + EM² = AM²

AE² + (AC/2)² = AM²

AE² + 24² = 576

AE² = 576 - 576/4 = 432

AE = √432 = 12√3 cm

Finally, we can find AG as:

AG = AE + 2GF = AE + 2(AD/4) (since G is the midpoint of EF)

AG = 12√3 + AD/2

But we know that AD = 2CE, so:

AG = 12√3 + CE

The value of CE that we previously discovered can therefore simply be added to obtain AG:

AG = 12√3 + CE = 12√3 + (AD/2) = 12√3 + (CE/2) = 12√3 + 6√3 = 18√3 cm.

Therefore, CE = AD/2 = (AG - 12√3)/2 = (18√3 - 12√3)/2 = 3√3 cm.

So, CE = 3√3 cm and AG = 18√3 cm.

6). AM is the middle of the right triangle ABC.

A line section connecting a triangle's vertex to the middle of the other side is the triangle's median. The median AM connects the right angle vertex A to the middle of the hypotenuse BC in the right triangle ABC.

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Two quadratic equations can share a solution if one quadratic equation has a positive discriminant and another quadratic equation has a discriminant equal to 0

What is meant by quadratic equation?

A quadratic equation is a polynomial equation of degree 2, which means it involves an unknown variable (usually denoted as x) that is raised to the power of 2.

If one quadratic equation has a positive discriminant, it means that the quadratic equation has two distinct real roots. On the other hand, if another quadratic equation has a discriminant equal to 0, it means that the quadratic equation has only one real root that is repeated.

Therefore, it is possible for two quadratic equations to share a solution if one quadratic equation has a positive discriminant and another quadratic equation has a discriminant equal to 0, but only if the repeated root of the second quadratic equation is one of the two roots of the first quadratic equation.

To illustrate this, let's consider the following two quadratic equations:

Quadratic equation 1: [tex]$x^2 - 4x + 3 = 0$[/tex]

Quadratic equation 2: [tex]$x^2 - 2x + 1 = 0$[/tex]

The discriminant of quadratic equation 1 is [tex]$b^2-4ac = 4 - 4(1)(3) = -8$[/tex], which is negative. Therefore, quadratic equation 1 has two distinct real roots, which can be found using the quadratic formula:

[tex]$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$[/tex]

Substituting the values of a, b, and c from quadratic equation 1, we get:

[tex]$x=\frac{4\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)}$[/tex]

[tex]$x=2\pm\sqrt{1}$[/tex]

[tex]x=3$ or $x=1$[/tex]

Therefore, the roots of quadratic equation 1 are [tex]x=3$ and $x=1$[/tex].

The discriminant of quadratic equation 2 is [tex]$b^2-4ac = 2^2 - 4(1)(1) = 0$[/tex]. Therefore, quadratic equation 2 has one repeated real root, which is:

[tex]$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$[/tex]

Substituting the values of a, b, and c from quadratic equation 2, we get:

[tex]$x=\frac{2\pm\sqrt{2^2-4(1)(1)}}{2(1)}$[/tex]

[tex]$x=1$[/tex] (repeated root)

As we can see, the repeated root of quadratic equation 2 is equal to one of the roots of quadratic equation 1, which is [tex]$x=1$[/tex]. Therefore, these two quadratic equations share a solution.

To learn more about quadratic equation visit:

https://brainly.com/question/1214333

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