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tamaranim1 [39]
3 years ago
9

Find all real numbers $s$ such that $-2(s - 7) > 4s + 8$. Give your answer in interval notation.

Mathematics
1 answer:
Stolb23 [73]3 years ago
5 0

Answer:

s\in (-\infty,1)

Step-by-step explanation:

Solve the inequality -2(s - 7) > 4s + 8

Use distributive property:

-2s+14>4s+8

Subtract 4s:

-2s+14-4s>4s+8-4s\\ \\-6s+14>8

Subtract 14:

-6s+14-14>8-14\\ \\-6s>-6

Divide by -6 (do not forget to change sign, because division by negative number changes the sign):

s

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Whats the answer to 8p squared minus 16 p=10
goblinko [34]
8p² - 16p = 10

8p² - 16p - 10 = 0      Divide through by 2

4p² - 8p - 5 = 0

Multiply first and last coefficients:  4*-5 = -20

We look for two numbers that multiply to give -20, and add to give -8

Those two numbers are 2 and -10.

Check:   2*-10 = -20         2 + -10 = -8

We replace the middle term of -8p in the quadratic expression with 2p -10p


 4p² - 8p - 5 = 0     

4p² + 2p - 10p - 5 = 0     

2p(2p + 1) - 5(2p + 1) = 0

(2p + 1)(2p - 5) = 0

2p + 1 = 0    or   2p + 5 = 0

2p = 0 -1              2p = 0 - 5

2p = -1                    2p = -5

p = -1/2                    p = -5/2

The solutions are p = -1/2  or  -5/2
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3 years ago
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Which value of a would make the following statement true -4(2x - a) = -8x + 16?
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2(9x^2+3)

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Step-by-step explanation:

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Answer:

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