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AURORKA [14]
2 years ago
12

Solve each inequality. (x – 3)(x - 2)^2 > 0

Mathematics
1 answer:
Rudiy272 years ago
6 0

Answer:

Write it like x>3 instead.

Work:

Solve for  x  by simplifying both sides of the inequality, then isolating the variable.

You might be interested in
What is the average rate of change between (3,2) and (1,9)
hjlf

Answer:

-0.25

Step-by-step explanation:

1-3= -2

9-2= 8

-2/8= -0.25

7 0
2 years ago
Help with volume all help is appreciated :)
Lina20 [59]

Answer:

It would be 0.41 ft^3

Step-by-step explanation:

Alright, to start, lets get the volume of the entire cinder block and the holes within it.

1.31 * 0.66 * 0.66 ---> 0.5706...

Next with the holes, they are both 0.33 wide, 0.39 long, and just as tall at 0.66 feet.

0.33 * 0.39 * 0.66 ---> 0.0849...

Since there's two of them: 0.1698...

To finish it, subtract the hole volume from the total volume;

0.5806 - 0.1698 = 0.4108

Rounds to <u>0.41 ft</u>^3

3 0
3 years ago
13.27= t - 24.45 show your work
Irina18 [472]
Subtract 24.45 from itself the subtract 13.27 by 24.45 and should t=37.73
7 0
3 years ago
Ab - bc+ ac, bc - ca+ ab, ca - ab-2bc
BabaBlast [244]

Answer:

ab-2bc+ca

Step-by-step explanation:

Complete question is given below

Add:-

ab-bc+ac,bc-ca+ab,ca-ab-2bc

We have to find the result after addition

Now, adding all expression

We get

ab-bc+ac+bc-ca+ab+ca-ab-2bc

Combine like terms then, we get

(ab+ab-ab)+(bc-bc-2bc)+(ac-ca+ca)

ab-2bc+ca

Hence,ab-bc+ac+bc-ca+ab+ca-ab-2bc=ab-2bc+ca

3 0
3 years ago
Four circles, each with a radius of 2 inches, are removed from a square. What is the remaining area of the square? (16 – 4π) in.
anyanavicka [17]
Since 4 circles are circumscribed by a square, then the side length of the square is 8

So the area of the square is 8^2 = 64

The area of 4 squares is 4 * pi*2^2 = 16pi

So the answer will be 64 - 16pi
6 0
3 years ago
Read 2 more answers
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