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3 years ago

14

What’s the total surface area for the shape?

1 answer:

max2010maxim [7]3 years ago

3 0

The answer is 420. Hope this helped.

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A certain recipe calls for 2 3/4 cups of sugar.if the recipe is to be doubled, how much sugar should be used

Morgarella [4.7K]

2 and 3/4 + 2 and 3/4 is 5 and 1/2.

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3 years ago

Sin(-x)= -cos x for all values of x. True or false

kogti [31]

Answer:

That's incorrect. The simplest way to show this is by evaluating the functions at a given point. Let's say x=0, then:

Sin(-x) = Sin(0) = 0

-cos x = -cos (0) = -1

Therefore, Sin(-x)≠-cos x.

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4 years ago

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if i have and 85 in my class and retook a test that i originally got a 72 what grade would i need for the average to be enough t

viktelen [127]

Assuming there is no specific percentages in the grade book, you would need a grade of 95 to get your whole grade to 90. You find this out by working backwards starting with 90 to find average. You take 90 times 2 which equals 180 and you subtract 85 from that to get 95! Brainliest?

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3 years ago

Find the Surface Area and Volume of Sphere with radius 12 cm.

Gnom [1K]

$Radius:r=12cm\\\\Surface\ area:\\A=4\pi r^2\to A=4\pi\cdot12^2=4\pi\cdot144=576\pi\ (cm^2)\\\\Volume:\\V=\frac{4}{3}\pi r^3\to V=\frac{4}{3}\pi\cdot12^3=\frac{4}{3}\pi\cdot1728=4\pi\cdot576=2304\pi\ (cm^3)$

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3 years ago

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Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago