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3 years ago

2(4a - 4b) + 4(3a + 2b)plz solve this!!!

Mathematics

1 answer:

ivann1987 [24]3 years ago

4 0

Answer:

2(4a - 4b) + 4(3a + 2b) = 20a

Step-by-step explanation:

2(4a - 4b) + 4(3a + 2b)

8a - 8b + 12a + 8b

8a + 12b - 8b + 8b

20a+ 0 = 20a

Thus, 2(4a - 4b) + 4(3a + 2b) = 20a

<u>-TheUnknownScientist</u>

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rosijanka [135]

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3 years ago

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Roman55 [17]

$(x^3+x^2+x+2) divide by (x^2-1)$

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3 years ago

Need help with AP CAL

anzhelika [568]

Answer: Choice C

$\displaystyle \frac{1}{2}\left(1 - \frac{1}{e^2}\right)$

============================================================

Explanation:

The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve $y = e^{-x}$

Think of the blue region as the floor of this weirdly shaped 3D room.

We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is $e^{-x}$ where 0 < x < 1

Let's compute the area of each general cross section.

$\text{area} = (\text{side})^2\\\\\text{area} = (e^{-x})^2\\\\\text{area} = e^{-2x}\\\\$

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.

This is what we want to compute

$\displaystyle \int_{0}^{1}e^{-2x}dx\\\\$

Apply a u-substitution

u = -2x

du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

Also, don't forget to change the limits of integration

If x = 0, then u = -2x = -2(0) = 0
If x = 1, then u = -2x = -2(1) = -2

This means,

$\displaystyle \int_{0}^{1}e^{-2x}dx = \int_{0}^{-2}e^{u}(-0.5du) = 0.5\int_{-2}^{0}e^{u}du\\\\\\$

I used the rule that $\displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$ which says swapping the limits of integration will have us swap the sign out front.

--------

Furthermore,

$\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]$

In short,

$\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]$

This points us to choice C as the final answer.

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2 years ago

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weqwewe [10]

SOLUTION

The question simply means that we should find the sum to infinity of the geometric series.

The formula of sum to infinity of a geometric serie is given by

$S_{\infty}=\frac{a}{1-r}$

Where

$\begin{gathered} S_{\infty}\text{ is the sum to infinity} \\ \\ a\text{ is the first term = 1} \\ \\ r\text{ is the common ratio = }\frac{2}{3} \end{gathered}$

So, this becomes

$\begin{gathered} S_{\infty}=\frac{a}{1-r} \\ \\ S_{\infty}=\frac{1}{1-\frac{2}{3}} \\ \\ S_{\infty}=\frac{1}{\frac{3-2}{3}} \\ \\ S_{\infty}=\frac{1}{\frac{1}{3}} \\ \\ S_{\infty}=3 \end{gathered}$

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1 year ago

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Nat2105 [25]

Answer:

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Step-by-step explanation:

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3 years ago

2(4a - 4b) + 4(3a + 2b)plz solve this!!! ​

2(4a - 4b) + 4(3a + 2b)plz solve this!!!