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3 years ago

Can someone help me with this thank you

Mathematics

1 answer:

nexus9112 [7]3 years ago

5 0

Answer:

324?

Step-by-step explanation:

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Evaluate M/N for M=10, N=-5, and P =-2

Elden [556K]

so it would be 10/-5 which would be -2

7 0

3 years ago

HELP MY PARENTS ARGUING ABOUT THIS ASSIGNMENT PLEASE DO THIS HURRY!!!

vredina [299]

I believe the answer are as follows: 1. A 2. B 3. A 4. D 5. B and 6. B

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2 years ago

Pleeeeeeeeaaaseeeeee help meeeeeeee!!!!!!!!!

zaharov [31]

Answer:

Option D.

$A =96\pi\ cm^2$

Step-by-step explanation:

The area of the circular bases is:

$A_c = 2\pi(a) ^ 2$

Where

$a=4\ cm$ is the radius of the circle

Then

$A = 2\pi(4) ^ 2$

$A = 32\pi\ cm^2$

The area of the rectangle is:

$A_r=b * 2\pi r$

Where

$b=8\ cm$

b is the width of the rectangle and $2\pi r$ is the length

Then the area of the rectangle is:

$A_r=8 * 2\pi (4)$

$A_r=64\pi\ cm^2$

Finally the total area is:

$A = A_c + A_r\\\\A = 32\pi\ cm^2 + 64\pi\ cm^2\\\\$

$A =96\pi\ cm^2$

7 0

3 years ago

Read 2 more answers

Evaluate using <br> Definite integrals

swat32

Since $[0,4]=[0,1]\cup(1,4]$ , we can rewrite the integral as

$\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt$

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

$\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt$

Both integrals are quite immediate: you only need to use the power rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}$

to get

$\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4$

Now we only need to evaluate the antiderivatives:

$\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15$

So, the final answer is 15.

4 0

3 years ago

{(-9,3) (5,-6) (3,-1), (-6,5), ?}Which ordered pair can replace the missing value and make the set if ordered pairs NOT a functi

Nookie1986 [14]

$\text{Slope}=\frac{-1--6}{3-5}=\frac{5}{-2}$ $\begin{gathered} y-y_1=\frac{5}{-2}(x-x_1) \\ \\ y--6=\frac{5}{-2}(x-5) \\ \\ -2(y+6)=5(x-5) \\ -2y-12=5x-25 \\ 5x+2y=25-12 \\ 5x+2y=13 \end{gathered}$

The missing values are (-6, -9) ( The last option)

Explanation:

x= - 6 has already produced the value of y as 5, the only pair in the given option the make the given ordered pairs NOT to be a function is (-6, -9) since it will give the value of y to be - 9.

4 0

1 year ago