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3 years ago

7

The area of a triangular block is 64 square inches. If the base of the triangle is twice the height, how long are the base and t

he height of the triangle? Will award brainliest!!!

1 answer:

jek_recluse [69]3 years ago

8 0

Algebra problem really.

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Find the area of the circle. (Take pi 22/7

Vikki [24]

Answer:

1386cm^2.

Step-by-step explanation:

radius of circle=21cm

area of circle(a)=?

we know,

A=pi r^2

=22/7×(21)^2

=1386cm^2

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2 years ago

On number 4, how do you figure out if it's an enlargement or a reduction?

aalyn [17]

A and C is enlargement
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3 years ago

Graph the quadratic function f(x) = -2x^2 + 5x + 2

larisa86 [58]

Answer:

Step-by-step explanation:

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3 years ago

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What is the solution to the equation? 7x + 6 = -22

trapecia [35]

C.) x = -4 is the correct answer

Step-by-step explanation:

In order to solve the equation we have to isolate x on one side of equation

Given

7x + 6 = -22

Subtracting 6 from both sides

$7x+6-6=-22-6\\7x=-28$

Dividing both sides by 7

$\frac{7x}{7}=\frac{-28}{7}\\x=-4$

Hence,

C.) x = -4 is the correct answer

Keywords: Linear equation

Learn more about linear equation at:

#LearnwithBrainly

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3 years ago

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note

Viktor [21]

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

By integrating both sides, we get

$\displaystyle-\ln(1-x)=C+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

If $x=0$ , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

$\displaystyle\ln(1-x)=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

We can shift the index to simplify the sum slightly.

$\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$ .

$\displaystyle x\ln(1-x)=-\sum_{n=1}^\infty\frac{x^{n+1}}n$

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

4 0

3 years ago