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3 years ago

9

A water sprinkler sprays in a circular pattern a distance of 10 ft from the sprinkler to the edge. What is the circumference of

the spray? (π = 3.14)

1 answer:

vfiekz [6]3 years ago

7 0

Answer:

<h2>62.8 ft</h2>

Step-by-step explanation:

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A(a-bx)/x=2/3<br> PLEASE

RSB [31]

Answer:

This cannot be solved but it can be simplified

first we open the bracket

so we have

a²- abx/x = 2/3

next we cross multiply so it will be 3(a² - abx) = 2x

we get

3a² - 3abx = 2x

Next we factorise so we can get

3a(a-bx) =2x

I dont know what exactly you where asked but I hop this helps

7 0

3 years ago

Whats the difference between 7 7/8 - 3 1/4

faust18 [17]

7 7/8 - 3 1/4

7 7/8 = 6.125

3 1/4 = 0.75

so 6.125 - 0.75

= 5.375

as a mixed fraction the answer is 5 3/8

4 0

3 years ago

What are the solutions of the quadratic equation (x + 3)(x +3) = 49? Ax = -2 and x = -16 Bx = 2 and x = -10 Cx = 4 and x = -10 D

Ivahew [28]

Answer:

4 and -10

Step-by-step explanation:

$\displaystyle (x + 3)(x +3) = 49 \\ x^2+3x+3x+9=49 \\ x^2 +6x+9=49 \\ x^2 + 6x + 9 - 49 = 0 \\x^2+6x-40=0 \\\\ \Delta=b^2-4ac \\ \Delta=6^2-4 \cdot 1 \cdot (-40) \\ \Delta=36+160 \\ \Delta=196 \\ \\ X_{1,2}=\frac{-b \pm \sqrt{\Delta} }{2a} \\ \\ X_1=\frac{-b+\sqrt{\Delta} }{2a} = \frac{-6+14}{2} = \frac{8}{2}=4 \\ \\ X_2=\frac{-b-\sqrt{\Delta} }{2a} = \frac{-6-14}{2}=\frac{-20}{2} = -10$

4 0

3 years ago

Approximate the change in the volume of a sphere when its radius changes from r = 40 ft to r equals 40.05 ft (Upper V (r )equal

alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere ( $V$ ), measured in cubic feet, is represented by the following formula:

$V = \frac{4\pi}{3}\cdot r^{3}$

Where $r$ is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

$\Delta V = \frac{\partial V}{\partial r}\Delta r$

The derivative of the volume as a function of radius is:

$\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}$

Then, the change in volume is expanded:

$\Delta V = 4\pi \cdot r^{2}\cdot \Delta r$

If $r = 40\,ft$ and $\Delta r = 40\,ft-40.05\,ft = 0.05\,ft$ , the change in the volume of the sphere is approximately:

$\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)$

$\Delta V \approx 1005.310\,ft^{3}$

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

7 0

3 years ago

A line has a y-intercept of 4 and a slope of 2/3. Explain how you can use this

Dafna1 [17]

Answer:

0,2 0r 2/2

Step-by-step explanation:

8 0

3 years ago

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