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2 years ago

Can someone do this quick! I need the volume!!

Mathematics

2 answers:

Dovator [93]2 years ago

4 0

Answer:

1125000

Step-by-step explanation:

just multiply them all

iren [92.7K]2 years ago

4 0

Answer:

i believe it should be 1125000

Step-by-step explanation:

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What is the product of -3 and 5 it's on engenuity so I need answer

Katarina [22]

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2 years ago

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Svetlanka [38]

The answer is option 2,

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3 years ago

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HELP ASAP f(x)=x² what is g(x)?

murzikaleks [220]

Just by comparing the plots of f(x) and g(x), it's clear that g(x) is just some positive scalar multiple of f(x), so that for some constant k, we have

g(x) = k • f(x) = kx² = (√k x)²

The plot of the transformed function g(x) = (√k x)² passes through the point (1, 4), which means

g(1) = (√k • 1)² = 4

and it follows that k = 4. So g(x) = 4x² = (2x)² and B is the correct choice.

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2 years ago

Find the area of a plane figure bounded by lines

natulia [17]

Answer: 4.5

Step-by-step explanation:

First, find the points of intersection by solving the system.

y = x² + 2x + 4

y = x + 6

Solve by substitution:

x² + 2x + 4 = x + 6 ⇒ x² + x - 2 = 0 ⇒ (x + 2)(x - 1) = 0 ⇒ x = -2, x = 1

Now, integrate from x = -2 to x = 1

$\int\limits^1_2 {(x+6)-(x^{2}+2x+4) } \,$ the bottom of the integral is -2

= $\int\limits^1_2 {x+6-x^{2}-2x-4 } \,$

= $\int\limits^1_2 {-x^{2}-x+2 } \,$

= $\frac{-x^{3}}{3} - \frac{x^{2}}{2}+2x\int\limits^1_2 {} \,$

= $(\frac{-1^{3}}{3} - \frac{1^{2}}{2}+2(1)) - (\frac{-(-2)^{3}}{3} - \frac{(-2)^{2}}{2}+2(-2))$

= $(\frac{-1}{3} - \frac{1}{2} +2) - (\frac{8}{3} -\frac{4}{2} -4)$

= $\frac{-9}{3} + \frac{3}{2} +6$

= -3 + 1.5 + 6

= 4.5

3 0

3 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

3 0

3 years ago

Can someone do this quick! I need the volume!!​

Can someone do this quick! I need the volume!!