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Andrew [12]
3 years ago
5

Could yalll PLS ANSWER ITS DUE LATE IVE BEEN ASKING FOREVER :(

Mathematics
2 answers:
OLga [1]3 years ago
8 0

Step-by-step explanation:

- 24 \div 4 =  - 6

[tex] - 24 \div 4 = \\ \frac{ - 24}{4} \\ = \frac{ - 12}{2} \\ = /frac{-6}{1}

Levart [38]3 years ago
5 0

Answer:

-6

-6/1 or 6/-1

Step-by-step explanation:

-24/4 is equal to -6 which is the same as -6/1 or 6/-1

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Can someone please help me on number 16-ABC
melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

-2x\:

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)>10\left(-1\right)

Simplify

2x>-10

Divide both sides by 2

\frac{2x}{2}>\frac{-10}{2}

x>-5

-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

-2x>-6

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)

Simplify

2x

Divide both sides by 2

\frac{2x}{2}

x

-6

Thus, the two intervals:

\left(-\infty \:,\:3\right)

\left(-5,\:\infty \:\right)

The intersection of these two intervals would be the solution to both inequalities.

\left(-\infty \:,\:3\right)  and \left(-5,\:\infty \:\right)

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR  -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0
3 years ago
I will mark you brainiest if you can answer this
Klio2033 [76]

Answer:

B

Step-by-step explanation:

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The answer is B i think
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Can someone help me solve for x?
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9x+72 = 4x+112
5x = 40
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8. (02.02 LC)
RSB [31]

Answer:

5 \sqrt{3}  - 20 \sqrt{2}

Step-by-step explanation:

simpify \\  \sqrt{75}  - 4 \sqrt{8}  - 3 \sqrt{32}  \\ from \: the \: above \: get \: two \: numbers \:  \\ such \: that \: when \: multiplied \: gives \\  \: the \: numbers \: inside \: the \: root \\ and \: one \: of \: the \: numers \: should \\ be \: a \: perfect \: square. \\  \sqrt{25 \times 3} - 4 \sqrt{4 \times 2}   -  3\sqrt{16 \times 2}  \\ 5 \sqrt{3}  - 4 \times 2 \sqrt{2}  -3 \times  4\sqrt{2}  \\ 5 \sqrt{3}  - 8 \sqrt{2}  - 12\sqrt{2}  \\ 5 \sqrt{3}  - 20 \sqrt{2}

3 0
2 years ago
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