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2 years ago

The circumference of a circle is 163.28 cm. What is the radius?

1 answer:

8 0

25.98681911 or rounded 26

C=pi(diameter)
163.28= pi(diameter)
Divide pi from both sides
51.97363822
Then divide by 2 to get radius

25.98681911

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The axis of symmetry for the function f(x) = –2x2 + 4x + 1 is the line x = 1. Where is the vertex of the function located?

egoroff_w [7]

Answer:

(1, 3)

Step-by-step explanation:

You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola. The first few steps are as follows. Set the parabola equal to 0 so you can solve for the vertex. Separate the x terms from the constant by moving the constant to the other side of the equals sign. The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out). Let's start there. The first 2 steps result in this polynomial:

$-2x^2+4x=-1$ . Now we factor out the -2:

$-2(x^2-2x)=-1$ . Now we complete the square. This process is to take half the linear term, square it, and add it to both sides. Our linear term is 2x. Half of 2 is 1, and 1 squared is 1. We add 1 into the set of parenthesis. But we actually added into the parenthesis is +1(-2). The -2 out front is a multiplier and we cannot ignore it. Adding in to both sides looks like this:

$-2(x^2-2x+1)=-1-2$ . Simplifying gives us this:

$-2(x^2-2x+1)=-3$

On the left we have created a perfect square binomial which reflects the h coordinate of the vertex. Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:

$-2(x-1)^2+3=y$

From this form,

$y=-a(x-h)^2+k$

you can determine the coordinates of the vertex to be (1, 3)

5 0

3 years ago

Read 2 more answers

How to solve this problem I don’t get it at all

Zarrin [17]

Use a calculator to help u and u will get your answer

8 0

3 years ago

Who had the quickest average time per mile?

Ugo [173]

Answer:

nina as faster

6 1/2 < 7 1/2

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2 years ago

Question 2 of 20

Mademuasel [1]

2+v12 is the answer

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3 years ago

Mathematical induction, prove the following two statements are true

adelina 88 [10]

Prove:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}$
____________________________________________

Base Step: For n=1:
$n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1$
and
$4-\dfrac{n+2}{2^{n-1}}=4-3=1$
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}$
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}$

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
$=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}$

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
$1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}$

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
$=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}$

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
$=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}$

Distribute the -2 and combine the fractions together,
$=4+\dfrac{-2k-4+(k+1)}{2^{k}}$

Combine like-terms,
$=4+\dfrac{-k-3}{2^{k}}$

pull the negative back out,
$=4-\dfrac{k+3}{2^{k}}$

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
$=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}$

3 0

3 years ago