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lianna [129]
3 years ago
5

The circle below is centered at the origin and has a radius of 5. What is its

Mathematics
1 answer:
nasty-shy [4]3 years ago
3 0

Answer:

Option B is the correct answer

Step-by-step explanation:

B. x2 + y2 = 25

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Help ill Mark brainlist :)
Marianna [84]

Answer:

The area would be 40.5 cm ^2

Step-by-step explanation:

Well first you don’t gotta worry about some isosceles trapezoid thingy.

You just gotta know the formula for a trapezoid is (b1+b2)/2)h

B1 in this case is or base 1 (9 or 4.5)

b2 in this case is or base 2 (9 or 4.5)

(you can only choose one base for each)

ANd you add those 2 up and you’d get 13.5

13.5/2 which is 6.75

Now we multiply it by the height which is 6 and gte 40.5

40.5cm ^2 is the final answer

6 0
3 years ago
HELP FAST I'LL MARK YOU BRAINLIEST What is an equation of the line that passes through the point (−2, 3) and is perpendicular to
igomit [66]

Answer: y = -2/3x + 5/3

Step-by-step explanation:

  • The slope of a perpendicular line is opposite reciprocal --> -2/3
  • Using point given (-2, 3), find b.
  • 3 = -2(-2/3) + b
  • 3 = 4/3 +b
  • 5/3 = b

5 0
3 years ago
You set a goal of creating a $16,000 emergency fund. You earn a salary of $40,000 per year
Lelechka [254]
D because it’s more longer than the other dates
7 0
3 years ago
Read 2 more answers
I NEED HELP PLEASE <br> ACHIEVE QUESTION
Mumz [18]

Answer:

its A

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Differentiate x^2 e^x logx
zimovet [89]

Product rule:

\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)

=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}

Power rule:

\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x

The exponential function is its own derivative:

\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x

Assuming the base of \log x is e, its derivative is

\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x

But if you mean a logarithm of arbitrary base b, we have

y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}

\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}

So we end up with

2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x

=xe^x(2\log x+x\log x+1)

8 0
3 years ago
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