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3 years ago

5

Find the volume of a cone. Either enter an exact answer in terms of or use 3.14 for pi and round your final answer to the neares

t hundredth. The radius is 2 the height is 9

1 answer:

IrinaK [193]3 years ago

5 0

Answer:

12 units³

Step-by-step explanation:

The required formula is V = (1/3)πr²h, where r is the radius and h is the height. Here we get:

V = (1/3)π(2)²(9) = (1/3)π(9)(4) = 12 units³

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Find the value of 24/25 divided by 4/5

Archy [21]

The answer to the question is 0.048

3 0

3 years ago

Which equation can be used to find the solution of (1/3)^d−5=81 ?

horsena [70]

Let's solve your equation step-by-step.

Step 1: Add 5 to both sides.

$\frac{1}{3}^{2} -5+5=81+5$

$\frac{1}{3}^d =86$

Step 2: Solve Exponent to get.

d = −4.054522 or -4.05 hope this helps

4 0

3 years ago

How do i turn -24/5 into a decimal answers plz thxs :)

kow [346]

Answer:

-4.8

Step-by-step explanation:

-24/5 = (-24) $\div$ 5

= -4.8

7 0

3 years ago

Read 2 more answers

ASNWER CORRECTLY AND YOU WILL GET 40 POINTS ALONG WITH BRAINLIEST

murzikaleks [220]

Answer: Angle ABC = 60, angle CBD = 120 and angle GFH = 60

Step-by-step explanation: Line ABD is parallel to line EFG. Line line CFH is a straight line that cuts across both parallel lines. Therefore, angle FBD and angle HFG are corresponding angles. That means angle FBD equals 3x. Also 3x plus 6X equals 180. That is,

3x + 6x = 180 {Sum of angles on a straight line equals 180}

9x = 180

Divide both sides of the equation by 9

x = 20.

That means angle 6x measures 6(20) and that is 120 degrees.

Also angle 3x measures 3(20) and that is 60 degrees.

Angle ABC + Angle CBD = 180 {Sum of angles on a straight line equals 180}

Angle ABC + 120 = 180

Angle ABC = 180 - 120

Angle ABC = 60

Also angle CBD equals 6x, and x = 20. Therefore angle CBD = 6 x 20

Angle CBD = 120.

And then, angle GFH = 3x, and x equals 20. Hence angle GFH = 60.

Therefore angle ABC = 60, angle CBD = 120 and angle GFH = 60.

7 0

3 years ago

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

Marrrta [24]

Given:

$f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6$

Required:

$We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.$

Explanation:

Given equation is

$F(x)=\sqrt[4]{f(x)g(x)}.$ $F(x)=(f(x)g(x))^{\frac{1}{4}}$ $F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}$

Differentiate the given equation for x.

$Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.$

$F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)$ $F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)$

Replace x=3 in the equation.

$F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)$ $Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}$ $F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)$ $F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-6-1}{4}$ $F^{\prime}(3)=\frac{-7}{4}$

Final answer:

$F^{\prime}(3)=\frac{-7}{4}$

8 0

1 year ago