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NemiM [27]
3 years ago
14

PLEASE HELP! I WILL MAKE YOU BRAINLIST!

Mathematics
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer: A.

Step-by-step explanation: yes

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points m and n are two vertices of a right triangle. they are the endpoints of the hypotenuse mn of the triangle. the third vert
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Step-by-step explanation:

4 0
3 years ago
Someone please help me thank you
german

Answer:

4

Step-by-step explanation:

7x = x² - 8

=> x² - 7x - 8 = 0

use quadratic formula:

a = 1, b = -7, c = -8

x = \frac{-(1)±\sqrt{(-7)^{2} - 4(1)(-8) } }{2(1)} <em>(pls ignore the "A" I don't know why it's showing up)</em>

=> x = \frac{-1±\sqrt{49 + 32} }{2}

=> x = \frac{-1±9 }{2}

=> x = \frac{-1 +9}{2} = 4 or \frac{-1-9}{2} = -5 <em>(the answer is only 4 since it's asking for the positive solution)</em>

6 0
3 years ago
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Heather is a cashier. She can ring up to 12 coustomers in 9 weeks. At this rate how many minutes does it take her to ring up 4 c
Solnce55 [7]
12 customers = 9 mintues

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Divide by 3 through
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12 ÷ 3 customers = 9 ÷ 3 minutes
4 customers = 3 minutes

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Answer: She can ring up 4 customers in 3 minutes
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3 0
4 years ago
The school science club raised $325.After buying equipment for an experiment they had $168 left.How much did they spend ?
Vsevolod [243]
325-x=168
325-168=x
x=157
325-157=168

They spent $157.
5 0
4 years ago
Find the absolute extrema of the function \(f(x)= xe^{- x^2/18}\) on the interval \([-2,4]\).
kenny6666 [7]
<span>Ans : f(x)=xe^(-x^2/8) differentiating by applying product rule, f'(x)=x(e^(-x^2/8))((-2x)/8)+e^(-x^2/8) f'(x)=e^(-x^2/8)(-x^2/4+1) f'(x)=-1/4e^(-x^2/8)(x^2-4) Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers. Now to find the critical numbers, solve for x for f'(x)=0. -1/4e^(-x^2/8)(x^2-4)=0 x^2-4=0 , x=+-2 e^(-x^2/8)=0 ,</span>
3 0
3 years ago
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