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Viefleur [7K]
3 years ago
7

Solve for x. Round to the tenths place. Can somebody help

Mathematics
1 answer:
VLD [36.1K]3 years ago
7 0

Answer:

12.12

Step-by-step explanation:

7^2+x^2=14^2, as by the Pythagorean theorem.

49+x^2=196,

x^2=147,

square root of 147;

it's close to 144,

thus, 12.12

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Sand will be placed under a circular pool with a diameter of 20 feet. The sand depth should be 2 inches. How many cubic feet of
JulsSmile [24]

you need 628 square feet of sand.

<h3>How many cubic feets of sand are needed?</h3>

Here we need to find the volume of the sand needed.

The volume filled with sand will be a cylinder of diameter of 20ft, and a height of 2 inches.

First, the radius is half of the diameter, then:

R = 20ft/2 =10ft

Remember that:

1ft= 12in

Then:

R = 10ft = 10*(12 in) = 120 in

And remember that the volume of a cylinder of radius R and height H is:

V  = pi*R^2*H

Where pi = 3.14

Then we have:

V = 3.14*(120 in)^2*2in = 90,432 in^2

To get that in feets, we need to divide by 12 squared, then:

90,432 in^2 = 90,432ft^2/(12^2)  = 628 ft^2

So you need 628 square feet of sand.

If you want to learn more about volumes:

brainly.com/question/1972490

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7 0
2 years ago
Simplify completely 12 times x to the third power minus 4 times x to the 2nd power plus 8 times x all over negative 2 times x. −
Nimfa-mama [501]
The answer is: −6x2 + 2x − 4 Solution: 12x^3 - 4x^2 + 8x ÷ -2x = -6x +2x - 4
6 0
4 years ago
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If adding two fractions that are both greater than one half what must be true about the sum
lys-0071 [83]

Answer:

The solution is greater than 1.

4 0
3 years ago
The length of a rectangle is five times the width the area is 196 square yards find the length and width of the rectangle
snow_lady [41]
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3 years ago
Find dy/dx for y=log (tan2x)
nekit [7.7K]

Answer:

\frac{dy}{dx} = \frac{2}{sin2x cos2x}

Step-by-step explanation:

y = log (tan 2x)

Applying chain rule:

\frac{dy}{dx} = \frac{d}{dx} (1st function) . \frac{d}{dx} (2nd function). \frac{d}{dx}(3rd function)

\frac{dy}{dx} = \frac{d}{dx} log(tan 2x) . \frac{d}{dx} tan2x. \frac{d}{dx} 2x

\frac{dy}{dx} = \frac{1}{tan2x} \times sec^{2}2x \times 2

\frac{dy}{dx} = \frac{2sec^{2}2x }{tan2x}

Now since sec^{2}x = \frac{1}{cos^{2}x } \ and\ tanx = \frac{sinx}{cosx}

∴ \frac{dy}{dx} = \frac{2cos2x }{sin2x cos^22x}

∴ \frac{dy}{dx} = \frac{2}{sin2x cos2x}

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