The sum of the roots of a quadratic is 1 and the product of the roots is -35/4.

a. find the quadratic.

b. find the roots

The height of the tree is approximately equal to the tangent of 768 radians multiplied by 50 meters, rounded to the nearest meter.

To find the height of the tree, we can use trigonometry. Let's assume that the height of the tree is represented by the variable "h."

In a right triangle formed by the forester, the base of the tree, and the top of the tree, the tangent of the angle is equal to the opposite side (height of the tree, h) divided by the adjacent side (distance from the base of the tree, 50 m).

Using the tangent function, we can write:

tan(angle) = [tex]\frac{h }{ 50}[/tex]

We can rearrange this equation to solve for h:

h = tan(angle) * 50

Now we can substitute the given angle into the equation and calculate the height:

h = tan(768) * 50

Calculating this in degrees might result in an error because the tangent function expects the angle to be in radians.

Therefore, we need to convert the angle to radians before evaluating the tangent.

There are 180 degrees in pi radians, so we can convert the angle as follows:

angle_radians = [tex]\frac{768 \* \pi }{ 180}[/tex]

Substituting this value into the equation:

h = tan(angle_radians) * 50

Now we can calculate the height using a calculator:

h ≈ [tex]tan(\frac{768 \pi }{ 180}) 50[/tex]

After evaluating this expression, we get the height of the tree. Rounding the answer to the nearest meter will give us the final result.

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Answer:

38

Step-by-step explanation:

Formula

Inscribed angle = Central angle/2

Here

Inscribed angle = x

Central angle = 76

x = 76/2

x = 38

B. Whole number enlarges, fraction reduces.

When a figure is dilated by a whole number, the image is enlarged by a factor of that whole number. For example, if a figure is dilated by a scale factor of 2, the image will be twice as large as the original.

On the other hand, when a figure is dilated by a fraction, the image is reduced. For example, if a figure is dilated by a scale factor of 1/2, the image will be half as large as the original.

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Sorry for bad handwriting

if i was helpful Brainliests my answer ^_^

Answer:

The answer is A

Step-by-step explanation:

Answer:

740

Step-by-step explanation:

37 times 20

Answer:

740 miles

Step-by-step explanation:

37 miles in 1 day

So we need to multiply 37*20 to find the number of miles for 20 days

So, he travels 740 miles

The greatest common what is the greatest common factor of 6a2b2 and 15a4b37 is option d.

To find the greatest common factor (GCF) of 6a^2b^2, 15a^4b^3, 7a^3b, and 3a^2b^2, follow these steps:

Step 1: Find the GCF of the numerical coefficients: The GCF of 6, 15, 7, and 3 is 1.

Step 2: Find the GCF of the 'a' terms: The lowest power of 'a' is a^2, so the GCF is a^2.

Step 3: Find the GCF of the 'b' terms: The lowest power of 'b' is b, so the GCF is b.

Combine the results from steps 1, 2, and 3: The GCF of 6a^2b^2, 15a^4b^3, 7a^3b, and 3a^2b^2 is 1a^2b.

Therefore, the GCF of 6a^2b^2 and 15a^4b^3 is 3a^2b^2, which is option (d).

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The conditions for inference are indeed met. The correct option is:
Yes, the conditions for inference are met.

The conditions for inference are met when conducting a confidence interval for a proportion if the following conditions are satisfied:

Random Sample: The sample should be a simple random sample or a random sample from a well-defined sampling frame. This ensures that the sample is representative of the population of interest.

Large Counts Condition: The sample size should be large enough so that both the number of successes (defective chips) and failures (non-defective chips) in the sample are at least 10. This ensures that the sampling distribution of the proportion is approximately normal.

Independence: The individual observations in the sample should be independent of each other.

In this scenario, the quality control specialist took a random sample of 100 chips from the day's production, which satisfies the random sample condition.

Now, let's check the Large Counts Condition.

The quality control specialist found 12 defective chips in the sample. To satisfy the Large Counts Condition, both the number of defective chips and the number of non-defective chips should be at least 10.

In this case, the number of defective chips is 12, and the number of non-defective chips is 100 - 12 = 88.

Both numbers are greater than 10, so the Large Counts Condition is met.

Since both the random sample condition and the Large Counts Condition are met, the conditions for inference are indeed met. Therefore, the answer is:

Yes, the conditions for inference are met.

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Answer:

12

Step-by-step explanation

60/100 to get the probability of success

0.6 * 20 attempts = 12

The distance between Town A and Town B is 550 km.

Let's denote the distance between Town A and Town B as D.

When the two cars first passed each other, let's assume that car A traveled a distance of x km and car B traveled a distance of D - x km.

Let's also denote the initial speeds of car A and car B as 6s and 5s, respectively, where s is some constant representing the speed of the slower car.

The time it took for the two cars to pass each other can be calculated using the formula:

time = distance / speed

For car A, the time it took to travel x km was:

x / (6s)

For car B, the time it took to travel D - x km was:

(D - x) / (5s)

Since the two cars traveled the same amount of time until they passed each other, we can set these two expressions equal to each other:

x / (6s) = (D - x) / (5s)

Solving for x, we get:

x = 6Ds / (11s)

After the speeds of both cars were reduced, car A's speed was (5/6) * 6s = 5s, and car B's speed was (3/4) * 5s = (15/4)s.

Let's denote the time it took for car A to travel the remaining distance from x to D as t.

Then, the time it took for car B to travel a distance of (D - x - 54) km is also t.

Using the new speeds, we can write the equation:

[tex](D - x - 54) = (15/4)s * t[/tex]

Solving for t, we get:

[tex]t = (4/15)(D - x - 54) / s[/tex]

The distance car A traveled after the two cars passed each other is:

D - x = D - 6Ds / (11s) = (5/11)D

The time it took for car A to travel this distance is:

[tex]t + x / (6s) = (4/15)(D - x - 54) / s + 6Ds / (66s)[/tex]

Setting these two expressions equal to each other and solving for D, we get:

D = 550 km

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To maximize Q = xy with the constraint x + y = 332, and given x = 4, we need to express the objective function in terms of y.

Since x = 4, we can rewrite the constraint as: 4 + y = 332

Now, solve for y:
y = 332 - 4
y = 328

Now, substitute the value of x into the objective function:
Q = (4)(y)

So, the objective function in terms of y is:
Q = 4y

To write the objective function in terms of y, we can solve for x in the constraint equation:

x + 332 = 4

x = 4 - 332

x = -328

Now we can substitute this value of x into the equation for Q:

Q = xy

Q = (-328)y

Q = -328y

Therefore, the objective function in terms of y is Q = -328y.

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Ans. Correct option in (c) 12!/11!

we know that

the formula of nPr is n! / (n−r)!.

in this question

n = 12 and r = 1

so putting values we get ,

= 12! / (12-1)!

=  12!/11!

The bearing of U from T in the image is 32 degrees South by West of T

In mathematics, bearings refer to the direction of an object or location in relation to two points. It is determined by the angle between the line joining them and that of the north.

Measured typically in degrees, it follows a cardinal system where 0° or 360° signifies North; East stands for 90°, South denotes at 180° while West represents 270°.

Thus, it can be seen that the bearing of U from T in the image is 32 degrees South by West of T

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The farmer would have made a profit of 5% if he had sold the goat for 2100 shillings.

Let's use the given terms and find out the percentage profit if the farmer had sold the goat for 2100 shillings.
Calculate the cost price of the goat
We know that the farmer made a loss of 28% by selling the goat for 1440 shillings. Let's represent the cost price as "CP".

We can write the equation:
[tex]CP \times  (1 - loss% ) = selling price (SP)[/tex]
[tex]CP \times  (1 - 0.28) = 1440[/tex]
Solve for CP
[tex]CP \times 0.72 = 1440[/tex]
CP = 1440 / 0.72
CP = 2000 shillings
Calculate the percentage profit
Now we want to find out the percentage profit if the farmer had sold the goat for 2100 shillings.

We can write the equation:
[tex](SP_{new - CP)} / CP \times  100 = profit[/tex]
[tex](2100 - 2000) / 2000 \times  100 = profit%[/tex]
[tex]100 / 2000 \times  100 = profit%[/tex]
5% = profit%.

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The perimeter is indeed 26 feet, which means they have used all the extra fencing.

Let's start by assuming that the length and width of the garden patch are whole numbers of feet, since we are asked to use only whole numbers.

Let's call the length of the garden patch "L" and the width "W".

We know that Liam has 26 feet of fencing left over. This fencing will be used to make the perimeter of the garden patch, which is given by:

Perimeter = 2L + 2W

We can substitute the value of the perimeter with the amount of fencing that Liam has:

26 = 2L + 2W

Simplifying this equation, we get:

13 = L + W

Since we want to use all the extra fencing, we know that the perimeter of the garden patch must be 26 feet. We can use this information to write another equation:

Perimeter = 2L + 2W = 26

We can substitute the value of 13 for L + W in this equation:

2L + 2W = 26

2L + 2(13-L) = 26

2L + 26 - 2L = 26

26 = 26

This equation is true, which means that our assumption that L and W are whole numbers is correct.

Therefore, the dimensions of the garden patch that Liam and his brother can make for their sister are 6 feet by 7 feet.

To check, we can calculate the perimeter:

Perimeter = 2L + 2W = 2(6) + 2(7) = 12 + 14 = 26

So the perimeter is indeed 26 feet, which means they have used all the extra fencing.

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Ali's Gulf club account will have a future value of BD 101.93 at the end of the term on October 7. The calculation was done using the formula for future value of an annuity with monthly payments, interest rate of 8% per year, and a term of 2.9 months.

We can first calculate the number of months from July 11 to October 7: 2 months and 27 days (or approximately 2.9 months).

Then, we can use the formula for future value of a present sum with simple interest

FV = P(1 + rt)

where FV is the future value, P is the present sum (in this case, BD 100), r is the annual interest rate (8% = 0.08), and t is the time in years (2.9/12 = 0.2417 years).

Substituting the values, we get

FV = 100(1 + 0.08*0.2417)

= 100(1.01934)

= BD 101.93

Therefore, the future value of Ali's golf club account is BD 101.93.

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The radius of the base of the cone is approximately 12.65 feet if The cone is formed from 3,200 ft of gravel.

Height of cone =  24 feet

The volume of the cone = [tex]3,200 ft^3[/tex]

To find the volume of the cone, the formula used here is:

V =π* [tex]r^2h[/tex]

Here, the values of V and H are known terms. we need to calculate the radius r of the cone.

π = 3.14 constant value

Substituting the values in the above equation, we get:

[tex]3,200 = (1/3)^2*(24)*[/tex] π

[tex]r^2[/tex]= 3,200 / (8π)

[tex]r^2[/tex] = 400 / π

r = 12.65

Therefore, we can conclude that the radius of the base of the cone is  12.65 feet.

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Answer:

B, A, C

Step-by-step explanation:

The rate is the another name for the slope.

A:

Change in y over the change in x.  You find the change by subtracting

[tex]\frac{7-3}{5-3}[/tex] = [tex]\frac{4}{2}[/tex] = 2

The rate is 2.

B:

Change in y over the change in x. You find the change by subtracting.

[tex]\frac{0-3}{-5-0}[/tex] = [tex]\frac{-3}{-5}[/tex] = [tex]\frac{3}{5}[/tex]

The rate is [tex]\frac{3}{5}[/tex].

C:

The rate is the number before the x in the equation.

The rate is 3.

Helping in the name of Jesus.

The problem presents a scenario where a square patio is placed in the centre of a 10m x 10m square backyard. The side length of the patio is given by x, and the remaining area of grass is expressed as A=-x^2+100.

To find the area of grass remaining, we can substitute different values of x into the equation.

For instance, if the patio is 5m x 5m, then x = 5 and the area of grass remaining is[tex]A = -5^2 + 100 = 75[/tex] square metres. Similarly, if the patio is 8m x 8m, then x = 8 and the area of grass remaining is [tex]A = -8^2 + 100 = 36[/tex] square metres.

As we can see, the area of grass remaining decreases as the size of the patio increases.

This problem illustrates the concept of content loaded and content remaining, where the initial content is the entire area of the square backyard, and the loaded content is the area of the square patio.

The remaining content is what is left after the loaded content is subtracted from the initial content. In this case, the loaded content is the patio area, and the remaining content is the grass area.

In summary, the area of grass remaining in the backyard after a square patio is placed in the centre can be calculated using the equation [tex]A=-x^2+100,[/tex] where x is the side length of the patio in metres.

The concept of content loaded and content remaining is also illustrated in this problem, where the loaded content is the patio area and the remaining content is the grass area.

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The surveyor is located about 239.6 feet away from the base of the Willis Tower.

The height of the Willis Tower is 165 feet.

The angle of elevation from the surveyor to the top of the antenna is about 3.41°.

The height of the antenna is about 135.9 feet.

Let's call the distance from the surveyor to the base of the Willis Tower "x", and let's call the height of the Willis Tower "h".

We can use trigonometry to solve for x and h. First, let's find x:

tan(34°) = h/x

x = h/tan(34°)

Now we can use the distance from the surveyor to the top of the building to solve for h:

h + 2595 = 2760

h = 165

So the height of the Willis Tower is 165 feet. Now we can solve for x:

x = 165/tan(34°) ≈ 239.6 feet

So the surveyor is located about 239.6 feet away from the base of the Willis Tower.

To find the angle of elevation from the surveyor to the top of the antenna, we can use trigonometry again:

tan(θ) = h/2760

θ = tan^(-1)(h/2760)

θ ≈ 3.41°

So the angle of elevation from the surveyor to the top of the antenna is about 3.41°.

Finally, we can use the height of the Willis Tower and the distance from the surveyor to the top of the antenna to solve for the height of the antenna:

tan(34°) = (h + a)/2760

a ≈ 135.9

So the height of the antenna is about 135.9 feet.

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The number of milligrams of sugar his spoon holds is 1250 milligrams.

To find out how many milligrams of sugar Chad's spoon holds, we first need to know how much sugar he removed in total. To do this, we can subtract the amount of sugar he needed (280 grams) from the amount he poured (295 grams).

295 grams - 280 grams = 15 grams

Next, we need to divide the total amount of sugar Chad removed (15 grams) by the number of spoonfuls he used (12).

15 grams ÷ 12 = 1.25 grams per spoonful

Finally, we can convert grams to milligrams by multiplying by 1000.

1.25 grams x 1000 = 1250 milligrams

Therefore, Chad's spoon holds 1250 milligrams of sugar.

It's important to note that when cooking or baking, precise measurements are crucial to the success of the recipe. Even small changes can greatly affect the outcome. While it's great that Chad was able to remove the excess sugar, it's best to be as accurate as possible from the start.

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The size of each section in a graph tells in relation to the portion of the data :

The magnitude of a graphic's segment indicates a specific attribute of the data being presented. In charts like pie charts or stacked bar graphs, each component's size denotes the relative frequency or proportion of data points in a particular category.

This implies that larger fragments reflect an increased number of data points for its associated categories whereas smaller ones represent categories with lesser data points.

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The statement "if point b is on ac and between points a and c, then mab + mbc = mac" describes the angle addition postulate in geometry.

In geometry, an angle is formed by two rays that share a common endpoint called a vertex. The measure of an angle is the amount of rotation between the two rays, usually measured in degrees or radians. The angle addition postulate states that if point B is on line segment AC and between points A and C, then the sum of the measures of angles MAB and MBC is equal to the measure of angle MAC. This postulate is used in various proofs and constructions in geometry, and it is also useful in real-world applications such as navigation, surveying, and engineering. The postulate is based on the fact that a straight angle measures 180 degrees, so if we know the measures of two angles that share a common ray, we can find the measure of the third angle by subtracting the sum of the first two angles from 180 degrees.

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Yes, Mark is correct he believes that doubling one of the dimensions would double the volume of his toolbox.

Volume refers to the space occupied by a 3-Dimensional space. The volume of a cuboid is given by:

V = l * b * h

where l is the length

b is the breadth

h is the height

V = 9 * 6 * 7

= 378 cubic inches

If we double any of the dimensions, like

By doubling the 9 we get

V = 18 * 6 * 7

= 756 cubic inches

By doubling the 6 we get

V = 9 * 12 * 7

= 756 cubic inches

By doubling the 7 we get

V = 9 * 6 * 14

= 756 cubic inches

Then the volume of the toolbox is doubled as shown above.

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The critical point is x = -1/2, the function has a local minimum at x = 1 and an absolute maximum at x = 4, and the absolute minimum is at x = -1/2.

a. The critical point is x = -1/2.

To find the critical point(s), we need to find where the derivative of the function is equal to zero or undefined. In this case, we have:

f(x) = -4x - x^2 + 2

f'(x) = -4 - 2x

Setting f'(x) equal to zero, we get:

-4 - 2x = 0

-2x = 4

x = -2/2

x = -1

However, we need to check if this value is in the given interval (1-4, 4). Since -1 is not in the interval, it is not a critical point.

Next, we check the endpoints of the interval.

When x = 1, f(x) = -4 - 1^2 + 2 = -3.

When x = 4, f(x) = -4 - 4^2 + 2 = -22.

So the function has a local minimum at x = 1, and an absolute maximum at x = 4, and no local maximum.

b. The local maximum is at x = 4, and the local minimum is at x = 1.

We can use the First Derivative Test to locate the local maximum and minimum points. If the derivative changes sign from positive to negative at a point, then it is a local maximum. If the derivative changes sign from negative to positive at a point, then it is a local minimum.

In this case, we have f'(x) = -4 - 2x. It is negative for x < -2 and positive for x > -2. Therefore, the function is decreasing for x < -2 and increasing for x > -2. Since the interval is (1-4, 4), the critical points are -2 and 4.

For x = 4, we have f'(4) = -4 - 2(4) = -12, which is negative, so x = 4 is a local maximum.

For x = 1, we have f'(1) = -4 - 2(1) = -6, which is negative, so x = 1 is a local minimum.

Therefore, the local maximum is at x = 4, and the local minimum is at x = 1.

c. The absolute maximum is at x = 4, and the absolute minimum is at x = -1/2.

To find the absolute maximum and minimum, we need to evaluate the function at the critical points and endpoints of the interval, and choose the largest and smallest values, respectively.

We have already found that the local maximum is at x = 4, and the local minimum is at x = 1. We also found that x = -1/2 is a critical point, but it is not in the given interval, so we can ignore it.

Evaluating the function at the endpoints of the interval, we get:

f(1) = -3

f(4) = -22

Therefore, the absolute maximum is at x = 4, and the absolute minimum is at x = 1/2.

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The clockwise rotation about the origin from the pre-image to the image​ include the following: a rotation of 90° clockwise.

In Mathematics, a rotation is a type of transformation which moves every point of the object through a number of degrees around a given point, which can either be clockwise or counterclockwise (anticlockwise) direction.

Additionally, the rotation of a point 90° about the center (origin) in a clockwise direction would produce a point that has these coordinates (y, -x).

Next, we would apply a rotation of 90° clockwise to the coordinate of triangle ABC as follows;

(x, y)                               →            (y, -x)

Coordinate A = (1, 1) → Coordinate A' = (1, -(1)) = (1, -1)

Coordinate B = (1, 4) → Coordinate B' = (4, -(1)) = (4, -1)

Coordinate C = (5, 1) → Coordinate C' = (1, -(5)) = (1, -5)

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Answer:

9, 10, 11, 12, 13.

Step-by-step explanation:

All possible values for the number of tables that the store must sell in order to meet the requirements are 9, 10, 11, 12, 13

The final values is (fog)(x) = f(g(x)) = f(0) = -8.

In the given problem, we are given two functions, f(x) and g(x). The function f(x) is a polynomial function, and g(x) is a constant function equal to 0. We are asked to find the composition of these two functions, that is, (fog)(x) and (gof)(x).

The composition of two functions f(x) and g(x) is denoted by (fog)(x) and is defined as follows:

(fog)(x) = f(g(x))

This means that we first evaluate g(x) and then use the output of g(x) as the input of f(x) to get the final output of (fog)(x).

In this case, since g(x) = 0, we have:

(fog)(x) = f(g(x)) = f(0)

To evaluate f(0), we substitute x = 0 in the expression for f(x):

[tex]f(x) = 2x^2 - 6x^2 + 4x - 8[/tex]

[tex]f(0) = 2(0)^2 - 6(0)^2 + 4(0) - 8[/tex]

f(0) = -8

Therefore, (fog)(x) = f(g(x)) = f(0) = -8.

Now, to find (gof)(x), we need to evaluate g(f(x)). Since f(x) is a polynomial function, we can find its value for any value of x. However, since g(x) is a constant function equal to 0, its output is always 0 for any input x. Therefore, g(f(x)) = 0 for all values of f(x).

This means that (gof)(x) = g(f(x)) = 0 for all x.

In summary, we found that (fog)(x) = -8 and (gof)(x) = 0.

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A nonspecific defense, also known as innate immunity, is the first line of defense that the body uses against pathogens, such as bacteria, viruses, and fungi.

This defense is considered nonspecific because it is not targeted towards a particular pathogen but rather provides a general defense mechanism that helps the body to prevent and fight infections. Examples of nonspecific defenses include physical barriers such as skin and mucous membranes, chemical barriers such as stomach acid and lysozyme, and cellular defenses such as phagocytic cells and natural killer cells.

2. The body's second line of defense is also part of the innate immune system and involves the activation of nonspecific leukocytes, including neutrophils, monocytes/macrophages, and natural killer cells. This defense takes effect when pathogens manage to breach the physical and chemical barriers of the body and enter into the tissues. These leukocytes then recognize and attack the invading pathogens through various mechanisms such as phagocytosis, release of cytotoxic substances, and activation of the complement system.

3. Nonspecific leukocytes play critical roles in the body's second line of defense by recognizing and attacking foreign pathogens. Neutrophils are the most abundant leukocytes in the body and are the first responders to infection. They are recruited to the site of infection and destroy pathogens through phagocytosis and release of antimicrobial substances. Monocytes/macrophages also play a crucial role in phagocytosis and release cytokines that activate other immune cells. Natural killer cells are responsible for recognizing and killing virus-infected cells and tumor cells.

Lastly, question 4. When Jera's skin became red and warm around the cut, it was a sign of inflammation, which is a normal response of the immune system to infection. Inflammation is characterized by redness, heat, swelling, and pain, and is caused by the release of chemicals such as histamine and cytokines in response to infection. The purpose of inflammation is to increase blood flow to the site of infection, bringing immune cells and nutrients to aid in the fight against the invading pathogens.

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