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Allushta [10]
3 years ago
13

Find the area of a circle with a diameter of 5 cm.

Mathematics
2 answers:
sergeinik [125]3 years ago
7 0

Answer:

19.635

Step-by-step explanation:

A=pie Radius^2

MariettaO [177]3 years ago
5 0

Answer:

6.25pi

Step-by-step explanation:

The area of a circle formula is A=pi(r²) and the diameter of the circle is 5 so divide that by 2 then square it to get 6.25 and then just put the pi after it

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90/200

Step-by-step:

Kahn Academy marked it correct

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3 years ago
Anyone knows how to do questions 7 and 8? 15 pts!!
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7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: 2x^2-4x+5. With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

2x^2+5=4x\\\\ 2x^2-4x+5=0

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)

Multiplying by (2-1) in the both sides:

(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ... Repeating the process, we obtain: ...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}
3 0
3 years ago
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Step-by-step explanation:

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2 years ago
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y  = a ( x - h)^2 + v             


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does not exist.
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