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ser-zykov [4K]
2 years ago
12

What’s the length of AZ. If A is between Y and Z, Y A = 22, AZ = 16x, and Y Z = 166

Mathematics
1 answer:
VLD [36.1K]2 years ago
7 0

Answer:

c) 144

Step-by-step explanation:

Since A is between Y and Z,

YZ = YA + AZ

116 = 22 + 16x\\16x = 116 - 22\\16x = 144\\x = 9

however, the question asks for the length of AZ, not the value of x.

Thus,

AZ = 16x

AZ = 16\\ × 9

AZ = 144

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Goryan [66]

558 gallons are used when the car has been driven 18 miles.

<u>Step-by-step explanation:</u>

<u>From the given data, it can be determined that :</u>

  • The number of gallons for 2 miles = 62 gallons
  • The number of gallons for 4 miles = 124 gallons
  • The number of gallons for 10 miles = 310 gallons

Let, the number of gallons for 18 miles = 'x' gallons

<u>To find x :</u>

compare the given data with the unknown data,

62 gallons / 2 miles = x gallons / 18 miles

keep x alone on one side,

x = (62×18) / 2

x = 31×18

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Therefore, 558 gallons are used to drive 18 miles.

3 0
3 years ago
Apply the distributive property to factor out the greatest common factor. 12+80=12+80
Ludmilka [50]

Answer: 92

Step-by-step explanation:

So 12 +80 = 4(3 + 20) = 92

5 0
2 years ago
PLEASE ANSWER: A camera usually sells for $68. Because it was slightly damaged, it sold for $59.90, what was the rate of the dis
Anestetic [448]
12% is the answer if u don’t need an explanation
3 0
1 year ago
Read 2 more answers
How can Kendra determine if the function is actually linear?
pashok25 [27]

Answer:

D. She can check to see if the rate of change between the first two ordered pairs is the same as the rate of change between the first and last ordered pairs.

Step-by-step explanation:

Find the rate of change between first two ordered pairs and the second two ordered pairs:

1. Points (2,4) and (3,9). Rate of change:

\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{9-4}{3-2}=\dfrac{5}{1}=5

2. Points (3,9) and (4,16). Rate of change:

\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{16-9}{4-3}=\dfrac{7}{1}=7

The rate of change for the linear function must the same for each two points on the graph of the function. In this case, the reate of change differs, so this function is not linear and correct option is D.

8 0
3 years ago
Read 2 more answers
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
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