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Nezavi [6.7K]
4 years ago
11

Should be easy for someone smart nit me LOL

Mathematics
1 answer:
Lina20 [59]4 years ago
7 0

They are equidistant from the origin. I.E., the mid-points are equidistant, or the same distance, to the center.

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PLEASE HELP!
kirza4 [7]
Equations with absolute value:

|f(x)| = k

Where k is a positive number; if k is a negative number, the equation is impossible (absolute value is always positive).

How to solve:

|f(x) | = k
- f(x) = k \: \: or \: \: f(x) = k

Then:
1. |x+7|=12
x+7=12 V -x-7=12
x=5 V -x=19
x=5 V x=-19
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2x+4=8 V -2x-4=8
2x=4 V -2x=12
x=2 V x=-6

3. 3|3k|=27
3×3k=27 V 3×(-3k)=27
9k=27 V -9k=27
k=3 V k=-3
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4. 5|b+8|=30
5×(b+8)=30 V 5×(-b-8)=30
5b+40=30 V -5b-40=30
5b=-10 V -5b=70
b=-2 V b=-14
{-14, -2}

5. |m+9|=5
m+9=5 V -m-9=5
m+9=5 V m+9=-5
4 0
3 years ago
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Bria took a test with 120 questions. She made an 85%. How many questions did Bria get wrong on the test?
Radda [10]

Answer:

you'd get eighteen questions wrong

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5x-2=3x+8 must find what X is equal to
serious [3.7K]
I had homework on this today lol

First, add 2 to both sides. It would then be 5x = 3x + 10. Then subtract 3x from both sides and you would end up with 2x = 10. Divide 2x by 2 and the 10 by 2. The answer is x = 5.
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Roman55 [17]
The domain is 2x and the range is -6
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How do you do this question?
Step2247 [10]

Answer:

D

Step-by-step explanation:

f(x)=(x-1)(x^2+2)^3\\f'(x)=(x-1)*3(x^2+2)^2*2x+(x^2+2)^3*1\\f'(x)=6x(x-1)(x^2+2)^2+(x^2+2)^3\\f'(x)=(x^2+2)^2[6x^2-6x+x^2+2]\\f'(x)=(x^2+2)^2(7x^2-6x+2)\\D

7 0
3 years ago
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