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Roman55 [17]
3 years ago
10

The area a if a circle is pi times the radius r squared

Mathematics
2 answers:
kirill [66]3 years ago
6 0
The area a if a circle is pi times the radius r squared
This is a true statement
Delicious77 [7]3 years ago
3 0

Totally, completely, and perfectly true.
Not a matter of opinion, and not open
to debate.

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A telemarketer earns $150 a week plus $2 for each call that results in sale. Last week she earned a total of $204. How many of h
cluponka [151]

Answer:

27 calls

Step-by-step explanation:

Let T(x) represent total sales.

Then T(x) = $150 + ($2/call)x, where x is the number of calls made.

If T(x)        = $204, we can solve for x, the number of calls made:

$204 = $150 + ($2/call)x, or

 $ 54

----------- = 27 calls

$2/call

8 0
3 years ago
Reflect the quadrilateral graph on the x axis graph <br><br>is the X-axis reflection right?​
agasfer [191]

Answer:

Step-by-step explanation:

No . B'  (reflection of B) and C' (reflection of C) are in the right position,  but D' should be at  (4, 1) and A' should be at (-2,-4).

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3 years ago
Assume that you are going to open a checking account. You are examining different banks and banking accounts, to choose a
Alexxandr [17]
Explain how the reactions of “the bloodless ghouls” to Orpheus’s song in paragraph 3 are important to the overall theme of the story. Cite evidence from the story in your response.
8 0
3 years ago
Your friend draws the diagram at the right
polet [3.4K]

Answer:

What he did wrong is that he did not flip all of the points. since S is 1 point up, it would then reflect 1 point down. point P is 3 up and the reflection should be 3 down. Q should be 3 down as well and R is 1 down below the x axis. The Y does not move for all of the points.

Step-by-step explanation:

I hope this helps!

8 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
4 years ago
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