Which equation is represented in the graph? parabola going down from the left and passing through the point negative 3 comma 0 then going to a minimum and then going up to the right through the points 0 comma negative 6 and 2 comma 0 Group of answer choices y = x2 − x − 6

Step-by-step explanation:

The area of a right triangle like this is   area = 1/2 * base * height

   base and height are the LEGS of the right triangle

  this one pictured piece area is:        1/2 * 14 in  * 11 in = 77 in^2

    He needs 12 of them ....so total is then   12 * 77 in^2 = 924 in^2 needed

The value of x is 29.5 units. Since the segments inside the circle have the same length, we can set  52.5  - x = 23 and get the value of x.

It is half of the diameter of the circle. The radius is used to calculate the area of a circle and the length of an arc or a circular sector.

In order to find the value of x, we need to first determine the radius of the circle.

In this problem, the first shorter side is 45 units, and the second shorter side is 27 units. Thus, we can find the radius of the circle by using the following method,

Radius of the circle = √(45² + 27²)

= √(2754) = 52.5 units

Now that we know the radius of the circle, we can use this information to find the value of x. We know that the total length of the first segment is  52.5 units. This means that the length of the segment outside the circle is x units and the length of the segment inside the circle is 52.5  - x units.

Similarly, the total length of the second segment is 50 units, so the length of the segment outside the circle is 27 units and the length of the segment inside the circle is

50 - 27 = 23 units.

Since the segments inside the circle have the same length, we can set  52.5  - x = 23 and solve for x.

52.5  - x = 23

x =  52.5  - 23

x = 29.5 units

Therefore, the value of x is 29.5 units.

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The bag contains $4.60 in coins.

What are coins?

Let's use P, N, D, and Q to represent the number of pennies, nickels, dimes, and quarters, respectively, in the bag.

We know that the total number of coins is 50, so:

P + N + D + Q = 50

We also know that 10% of the coins are pennies, so:

P = 0.1(50)

P = 5

There are 2 more nickels than pennies, so:

N = P + 2

N = 5 + 2

N = 7

Now we can use this information to find the number of dimes and quarters. Since there are 50 coins in total, we can substitute the values we have found for P and N into the first equation:

P + N + D + Q = 50

5 + 7 + D + Q = 50

12 + D + Q = 50

D + Q = 38

We also know the values of P and N, so we can find the total value of all the coins in the bag:

Total value = (value of pennies) + (value of nickels) + (value of dimes) + (value of quarters)

Total value = (5 cents x 5) + (7 cents x 5) + (10 cents x D) + (25 cents x Q)

Total value = 25 + 35 + 10D + 25Q

We can substitute D + Q = 38 into this equation to get:

Total value = 25 + 35 + 10(D + Q)

Total value = 60 + 10(38)

Total value = 460 cents, or $4.60

Therefore, the bag contains $4.60 in coins.

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The answer is y = ±1, as well as two complex solutions. The answer y = 3 is not a solution to the equation [tex]y^3 = 1.[/tex]

An equation is a statement that asserts the equality of two expressions. It typically contains one or more variables, which are symbols that represent unknown or varying quantities. The expressions on either side of the equals sign can include numbers, constants, functions, and other mathematical operations.

To solve[tex]y^3 = 1[/tex], we can take the cube root of both sides of the equation:

y = ∛1

There are three cube roots of 1, which are 1, -1/2 + √3/2 i, and -1/2 - √3/2 i, where i is the imaginary unit. These three roots form a complex conjugate pair.

Therefore, the solutions to the equation [tex]y^3 = 1[/tex] are:

y = 1

y = -1/2 + √3/2 i

y = -1/2 - √3/2 i

So the answer is y = ±1, as well as two complex solutions. The answer y = 3 is not a solution to the equation [tex]y^3 = 1.[/tex]

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Answer:

A

Step-by-step explanation:

Given:

r (radius) = 5 in

h (height) = 5 in

Find: A (surface area) - ?

.

[tex]a(surface) = a(lateral) + a(2 \: bases)[/tex]

First, we have to find the lateral surface:

[tex]a(lateral) = 2\pi \times r \times h = 2 \times 3.14 \times 5 \times 5 = 157 \: {in}^{2} [/tex]

Then, we have to find the area of the base and multiply it by 2, since the shown figure is a cylinder, which has 2 identical bases:

[tex]a(2 \: bases) = 2 \times \pi {r}^{2} = 2 \times 3.14 \times {5}^{2} = 157 \: {in}^{2} [/tex]

Now, let's add everything together and we'll get the answer:

[tex]a(surface) = 157 + 157 = 314 \: {in}^{2} [/tex]

Answer:

b

Step-by-step explanation:

just add then then after you add 27=27 multiply it by 2

Hence, point P's coordinates are (6, 3) as the midpoint of the line segment from R to Q, which is the precise halfway point between R and Q.

A point's location on a line or in space can be determined using a set of numbers called coordinates. An x-coordinate and a como, which denote the horizontal as well as vertical ranges from a bench mark known as the origin, are the two numbers that make up coordinates in two-dimensional space. Three numbers, the x, y, and z coordinates, which describe the distances from origin in the x, y, and g directions, respectively, might make up coordinates in three-dimensional space. Among other disciplines, coordinates are frequently employed in physics, arithmetic, and geometry.

given

The midpoint of the line segment from R to Q, which is the precise halfway point between R and Q, must be located in order to determine the coordinates of point P.

We can separately aver the x- and y-coordinates to get the middle.

The midpoint's x-coordinate is (4 + 8)/2, which equals 6.

(0 + 6)/2 = 3 is the midpoint's y-coordinate.

Hence, point P's coordinates are (6, 3) as the midpoint of the line segment from R to Q, which is the precise halfway point between R and Q.

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Answer:

(-3.5 , 2.3)

Step-by-step explanation:

I know

A number line that represent the values of x is shown in the graph attached below.

In Mathematics and Geometry, a number line simply refers to a type of graph with a graduated straight line which comprises both positive and negative numbers that are placed at equal intervals along its length.

This ultimately implies that, a number line primarily increases in numerical value towards the right from zero (0) and decreases in numerical value towards the left from zero (0):

x < - 1 or x > 1

-1 > x > 1

-3 < x ≤ 1

-5 ≤ x ≤ 0

In this scenario and exercise, we would use an online graphing calculator to plot each of the inequality as shown in the graph (number line) attached below.

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Answer:

11/3

Step-by-step explanation:

2[tex]\frac{1}{5}[/tex] = 11/5

2[tex]\frac{1}{5}[/tex] ÷ [tex]\frac{3}{5}[/tex] = 11/5 · 5/3 = 55/15 = 11/3

Answer:

36.4

Step-by-step explanation:

In morning, builder worked for = 25 in a hour.

Let hour be ? hr.

then 25? = Time he worked in morning.

In afternoon,he worked for= 910 of ? hour = 910? hr

Times he worked more = 910?/25? = 36.4 times

25*6 =

The dilated triangle has vertices

The coordinates of the vertex of the triangle before dillation is given as

P (-6, 4 )

Q (2, 8)

R (2, -4)

To dilate a figure by a scale factor of 1/2, we need to multiply the coordinates of each point by 1/2.

The coordinates of P are (-6, 4). Multiplying by 1/2, we get:

(-6 * 1/2, 4 * 1/2) = (-3, 2)

The coordinates of Q are (2, 8). Multiplying by 1/2, we get:

(2 * 1/2, 8 * 1/2) = (1, 4)

The coordinates of R are (2, -4). Multiplying by 1/2, we get:

(2 * 1/2, -4 * 1/2) = (1, -2)

Therefore, the dilated triangle has vertices P'(-3, 2), Q'(1, 4), and R'(1, -2).

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The selling price of the article without VAT is Rs. 6,429.13.

What is Algebraic expression ?

In mathematics, an algebraic expression is a combination of variables, constants, and mathematical operations, such as addition, subtraction, multiplication, and division, that represents a quantity or a relationship between quantities.

Let's assume the rate of discount to be 'x' and the rate of VAT to be 'y'.

According to the problem, the marked price of the article is Rs. 8,000.

After applying the discount, the selling price becomes:

Selling price = Marked price - Discount

Selling price = 8000 - (x÷100) * 8000

Now, we add the VAT to the selling price:

Selling price with VAT = Selling price + (y÷100) * Selling price

Selling price with VAT = [8000 - (x÷100) * 8000] + [(y÷100) * (8000 - (x÷100) * 8000)]

Selling price with VAT = 6689.60 (Given)

We are also given that the rate of discount was double the rate of VAT, i.e.,

x = 2y

Substituting this value in the above equation, we get:

6689.60 = [8000 - (2y÷100) * 8000] + [(y÷100) * (8000 - (2y÷100) * 8000)]

Simplifying this equation, we get:

6689.60 = 8000 * [1 - (2y÷100) + (y÷100) - (2y÷100) * (y÷100)]

6689.60 = 8000 * [1 - (3y÷50) + (y*y÷5000)]

Solving for y, we get:

y = 4

Now, we can calculate the selling price without VAT as follows:

Selling price without VAT = Selling price - (y÷100) * Selling price

Selling price without VAT = 6689.60 - (4÷100) * 6689.60

Selling price without VAT = Rs. 6,429.13

Therefore, the selling price of the article without VAT is Rs. 6,429.13.

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Answer:

Step-by-step explanation:

Search up the area of both. rectangle is LxW therefore the area is 18.  the other is trapezium therefore, ((a+b)xh)/2 therefore the area is 20. trapezium has larger area

Answer trapezoid has  greater area

Step-by-step explanation: rec: 6*3= 8,  trapezoid [ (6+2) * 5 ]1/2  = 20

Increasing [tex]\frac{15}{16}[/tex] by [tex]\frac{3}{4}[/tex] of it and then increasing the result by [tex]\frac{3}{5}[/tex] of it gives us the fraction of [tex]\frac{45}{32}[/tex].

To increase [tex]\frac{15}{16}[/tex] by [tex]\frac{3}{4}[/tex] of it, we can multiply [tex]\frac{15}{16}[/tex] by [tex]\frac{3}{4}[/tex], which gives us:

[tex]\frac{15}{16}[/tex] × [tex]\frac{3}{4}[/tex] = [tex]\frac{45}{64}[/tex]

Adding this result to [tex]\frac{15}{16}[/tex], we get:

[tex]\frac{15}{16}[/tex] + [tex]\frac{45}{64}[/tex] = [tex]\frac{225}{256}[/tex]

Now, to increase this result by [tex]\frac{3}{5}[/tex] of it, we can multiply [tex]\frac{225}{256}[/tex] by 3/5, which gives us:

[tex]\frac{225}{256}[/tex] × [tex]\frac{3}{5}[/tex] = [tex]\frac{135}{256}[/tex]

Adding this result to [tex]\frac{225}{256}[/tex], we get the final answer:

[tex]\frac{225}{256}[/tex] + [tex]\frac{135}{256}[/tex] = [tex]\frac{360}{256}[/tex]

Simplifying this fraction, we get:

[tex]\frac{360}{256}[/tex] = [tex]\frac{45}{32}[/tex]

The problem involves increasing a fraction by a certain percentage twice. To solve the problem, we first find the increase in the fraction by multiplying it with the given percentage. We then add the result to the original fraction to get the new fraction. This process is repeated for the second increase. Finally, we simplify the resulting fraction if possible. In this specific problem, we are increasing [tex]\frac{15}{16}[/tex] by [tex]\frac{3}{4}[/tex] of it and then increasing the result by [tex]\frac{3}{5}[/tex] of it to find the resulting fraction [tex]\frac{45}{32}[/tex].

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Answer:

Option D is the correct answer

D. f(n) = 95 + 60(n - 1)

Step-by-step explanation:

Solution:

Based on the above, the width of the rectangle is 10.

Let's use "w" to represent the width and "l" to represent the length of the rectangle.

From the problem, we know that the area of the rectangle is 40, so we can write:

Area = length × width

40 = l × w

We also know that the width is 2 less than 3 times the length, so we can write:

width = 3l - 2

Now we can substitute this expression for "w" in the equation for the area: 40 = l × (3l - 2)

Expanding the right side gives: 40 = 3l² - 2l

Rearranging and dividing by 2 gives:

3l²- 2l - 40 = 0

(3l + 10)(l - 4) = 0

Therefore, l = 4 (because the length cannot be negative), and we can find the width using the expression we found earlier:

width = 3l - 2

width = 3(4) - 2

width = 10

So the width of the rectangle is 10. Therefore, the answer is  10.

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x² + y² + 4x + 6y - 3 = 0  equation represents the general form a circle with a center at (â€"2, â€"3) and a diameter of 8 units.

The equation of a circle with center (h, k) and radius r can be written in the form (x - h)² + (y - k)² = r².

Given a center of (-2, -3) and a diameter of 8 units, first find the radius, which is half of the diameter:
r = 8 / 2 = 4
Now, substitute the center coordinates and radius into the equation:
(x - (-2))² + (y - (-3))² = 4²
(x + 2)² + (y + 3)² = 16
Now, expand the equation to get the general form:
(x² + 4x + 4) + (y² + 6y + 9) = 16
x² + 4x + y² + 6y + 4 + 9 - 16 = 0
Combine like terms to get the final equation:
x² + y² + 4x + 6y - 3 = 0

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Answer:

A random number list could be used to simulate the probability of rain each day this week. For example, we could assign the numbers 1-60 to the days where rain is expected and the numbers 61-100 to the days where it is not expected. Then, we could use a random number generator to simulate each day and count how many days it takes until it rains for the first time. By repeating this simulation many times and taking the average, we could estimate the experimental probability of it taking at least three days before it rains.

0 

A: 75% of items are classified as C items, 18.75% as B items, and 6.25% as A items.

A-B-C examination is an instrument utilized for stock administration, which orders things into three classifications: A, B, and C, in light of their worth and significance. Classification A things have a high worth however a low volume, and their administration requires close consideration and control. Class B things have moderate worth and volume and require moderate control. Classification C things have a low worth yet a high volume, and their administration can be loose. The reason for the examination is to recognize which things need the most consideration and control, and which things can be dealt with less consideration and cost-actually. The examination assists in streamlining with reviewing levels, decreasing expenses, and expanding benefits.

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Part A: Sample space of randomly selecting 2 lollipops with replacement:

{GG, GC, GL, CG, CC, CL, LG, LC, LL}

Part B: Sample space of randomly selecting 2 lollipops without replacement:

{GC, GL, CG, CL, LG, LC}

The experiment of randomly selecting 2 lollipops without replacement shows dependent events because the probability of drawing the second lollipop depends on what the first lollipop was.

For example, if the first lollipop drawn is grape, then the probability of drawing another grape lollipop is decreased because there is only one left in the bag.

The experiment of randomly selecting 2 lollipops with replacement shows independent events because each lollipop can be chosen without affecting the probability of choosing the other lollipops.

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Answer:

let n be the number

6 less than the number

= (n-6)^5 = 32

(n-6)^5 = 2^5

since powers are equal we equate the bases

n-6 =2

n=8

Answer: B= (5z+9) (5z-9)

Step-by-step explanation:

Answer:

Step-by-step explanation:

Step 1: Use the sum-product pattern

25z² - 81 =

25z² + 45z - 45z - 81 =

Step 2: Common factor from the two pairs

(25z² + 45z) + (−45z − 81) =

Step 3: Rewrite in Factored Form

5z (5z + 9) − 9 (5z + 9) =

(5 - 9) (5 + 9)

Solution:

B: (5 + 9) (5 - 9)

The true statements are:

C) -5 is a solution to f(x).

E) f(x) = 2x^2 + 9x - 5

F) 0 is a solution to f(x).

I) 5 is a solution to f(x).

K) -3 is a solution to f(x).

L) (-5,0) is an x-intercept of f(x).

From the given factors of f(x), we know that:

f(x) = 3 * (2x-1) * (x+5)

Using the  information, we can calculate  the  statements are true:

A) 1 is a solution to f(x).

Not necessarily true. We need to substitute x=1 into the expression for f(x) to check if it equals zero.

B) f(x) = 6x^2 - 33x - 15

Not true. The correct expression for f(x) is given above.

C) -5 is a solution to f(x).

True. Substituting x=-5 into f(x) gives us: f(-5) = 3*(-11)*0 = 0.

D) 3 is a solution to f(x).

Not necessarily true. We need to substitute x=3 into the expression for f(x) to check if it equals zero.

E) f(x) = 2x^2 + 9x - 5

True. This is the expanded form of f(x) using the given factors.

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The events ranked from least likely to most likely are:

Probability refers to potential. A random event's occurrence is the subject of this area of mathematics. The range of the value is 0 to 1. Mathematics has incorporated probability to forecast the likelihood of various events.

To rank the events from least likely to most likely, we need to consider the probability of each event occurring.

Event 3: Choosing a blue pen from Box A

This event is the least likely because there are no blue pens in Box A. The probability of choosing a blue pen is 0.

Event 2: Choosing a white pen from Box B

This event is more likely than Event 3 because there are white pens in Box B. The probability of choosing a white pen from Box B is 2/8 or 1/4.

Event 4: Choosing a green pen from Box A

This event is more likely than Event 2 because there are green pens in Box A. The probability of choosing a green pen from Box A is 3/15 or 1/5.

Event 1: Choosing a green or white pen from Box B

This event is the most likely because it includes both a green and a white pen in Box B. The probability of choosing a green or white pen from Box B is 6/8 or 3/4.

Therefore, the events ranked from least likely to most likely are:

Event 3: Choosing a blue pen from Box A

Event 2: Choosing a white pen from Box B

Event 4: Choosing a green pen from Box A

Event 1: Choosing a green or white pen from Box B

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The proportion of Year 9 pupils in the music club is approximately 46.7%.

A ratio or value that may be stated as a fraction of 100 is called a percentage.

To find the proportion of Year 9 pupils in the music club as a percentage, follow these steps:

To get proportion as a percentage divied number of Year 9 pupils by total pupils multiplied by 100.

Proportion as a percentage = [tex]\frac{Year \:9 \:pupils}{total \:pupils} \times 100[/tex]
Divide the number of Year 9 pupils by the total number of pupils.

78 (Year 9 pupils) ÷ 167 (total pupils)

= 0.4671
Multiply the result by 100 to convert the proportion to a percentage.
0.4671 × 100

= 46.71%.

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a) The probability that the battery lasts more than 4 hours is 0.758 (rounded to 3 decimal places).

b) The 25% value is 240 minutes, and the 75% value is 310 minutes.

c) The value of life in minutes that is exceeded with 95% probability is 357 minutes.

a) To find the probability that the battery lasts more than 4 hours (240 minutes), we first need to convert 240 minutes into a z-score.
z = (X - μ) / σ
z = (240 - 275) / 50
z = -0.7
Now, we'll use a z-table to find the probability that the battery lasts more than 4 hours:
P(Z > -0.7) = 1 - P(Z ≤ -0.7) = 1 - 0.2420 = 0.758
The probability that the battery lasts more than 4 hours is 0.758 (rounded to 3 decimal places).
b) To find the quartiles, we'll use the z-table to find the z-scores corresponding to 25% and 75%:
25%: z = -0.674
75%: z = 0.674
Now, we'll convert the z-scores back into minutes:
Q1 = μ + z * σ = 275 + (-0.674) * 50 = 240 (rounded to the nearest integer)
Q3 = μ + z * σ = 275 + (0.674) * 50 = 310 (rounded to the nearest integer)
The 25% value is 240 minutes, and the 75% value is 310 minutes.
c) To find the value of life in minutes that is exceeded with 95% probability, we first find the z-score corresponding to 95%:
95%: z = 1.645
Now, we'll convert the z-score back into minutes:
X = μ + z * σ = 275 + (1.645) * 50 = 357 (rounded to the nearest integer).

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Given:

[tex]\angle\text{L}=90^\circ[/tex]

[tex]\text{KJ}=41[/tex]

 [tex]\text{JL}=40[/tex]

  [tex]\text{LK}=9[/tex]

To find the cosine of angle J

By using cosine ratio,

[tex]\text{cos J}=\dfrac{\text{adjacent side}}{\text{hypotenuse}}[/tex]

       [tex]\text{cos J}=\dfrac{\text{JL}}{\text{JK}}[/tex]

         [tex]\text{cos J}=\dfrac{\text{40}}{\text{41}}[/tex]

The ratio of cosine of angle J is 40/41.

Answer:

blank1: 9

blank2: 30°

blank3: 60°

blank4: 90°

blank5: half

Step-by-step explanation:

The 30°-60°-90° triangle has some great shortcuts associated with it (there are math reasons, trig reasons for these shortcuts)

One of these great shortcuts is that the short leg is half the hypotenuse (or the hypotenuse is double the short leg)

For your question, the kite string, 18ft, is the hypotenuse. The kite height is half that--9ft.

see image.

The volume of cat food in the can is approximately 9.63 cubic inches.

What is the volume of a cylinder?

The volume of a cylinder can be found using the formula:

V = πr²h

where V is the volume, r is the radius of the circular base, h is the height of the cylinder, and π is the constant pi (approximately equal to 3.14).

To find the volume of the can of cat food, we can use the formula for the volume of a cylinder, which is V = πr²h, where r is the radius (half the diameter) and h is the height.

The radius of the can is 3.5/2 = 1.75 inches, and the height is 1 inch. So, the volume of the can is:

V = π(1.75)²(1)

= 9.62 cubic inches

≈ 9.63 cubic inches (rounded to the nearest hundredth)

Therefore, the volume of cat food in the can is approximately 9.63 cubic inches.

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