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3 years ago

Can I get help please

Mathematics

1 answer:

Masja [62]3 years ago

4 0

Answer:

A) 41.9

Step-by-step explanation:

The formula I use is π(pi)r(radius)^2h(height)/3, so it looks like πr^2(h/3) and I used that and I got 41.88 which rounds to 41.9 or A

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Giving brainliest, please answer as soon as possible.

Serjik [45]

Answer:

Step-by-step explanation:

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Each going from side midpoint to opposite vertex.

here is a drawing of all of them filled in:

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2 years ago

How much money is 2/3 of 24.00$ draw a diagram

Pachacha [2.7K]

Answer:

16.00$

Step-by-step explanation:

divide 24 dollars in three sections which is 8 dollars a section. 2/3 is 2 out of 3 sections(8+8) which is sixteen.

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2 years ago

Please help. I don't understand anything.

alina1380 [7]

Answer:

Step-by-step explanation:

The area of figure=Area of rectangle+Area of triangle

Area of figure=(x+2)(x-3)+1/2*(x)*(x+2)

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7 0

2 years ago

Read 2 more answers

Answer this real-world question

Mrrafil [7]

Answer:

After 15 days they'll both have $10

Explanation:

100 (Xavier's total)- 6x (6 dollars every day) = 160 (Zach's total amount) -10x (10 dollars every day)

$100-6x=160-10x\\$

+10x on both sides

$100+4x=160$

-100 on both sides

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/4 on both sides

$x=15$

4 0

2 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago