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2 years ago

Which inequality has -12 in its solution set?

Mathematics

1 answer:

enot [183]2 years ago

4 0

Answer:

D) $x+5\leq -4$

Step-by-step explanation:

We solve each of the inequalities

Option A

$x+6$

Option B

$x+4\geq -6$

$x\geq -6-4\\x\geq-10$

Option C

$x-3>-10\\x>-10+3\\x>-7$

Option D

$x+5\leq -4$

$x\leq -4-5\\x\leq -9$

Therefore, only option D has -12 in its solution set.

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If 5x – 4 = 31, what is the value of x?

Lunna [17]

Answer:

x = 5 $\frac{2}{5}$

Step-by-step explanation:

Hi there!

So, to answer this problem, I'll use the reverse method.

31 - 4 = 27,

So how many times, can 5 go into 27?

The closest factor of 5 to 27 is 25.

5 can go in to 27, 5 times.

Continuing on, there is a 2 left over, so the answer to your question is

x = 5 $\frac{2}{5}$

5 · 5 $\frac{2}{5}$ - 4 = 31

Let me know if I am incorrect and I will attempt to find the right answer for you.

<em>waffletowne</em>

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2 years ago

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Natalija [7]

C. Median because there is an outlier

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2 years ago

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What is 14/52 as a decimal

Kisachek [45]

Answer:

0.2692307692

Step-by-step explanation:

14/52 = 0.2692307692

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2 years ago

4. 7 + 14x<br> What does that mean

kotegsom [21]

7 + 14x means that a number x (x could be any real number) times 14 is added to 7.

So if x = 2:

7 + 14x = 7 + 14(2) = 28 + 7 = 35

Happy to help!

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2 years ago

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For what value of k will the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent?

konstantin123 [22]

Answer:

After solve the equations we get value of k=3

Step-by-step explanation:

We need to find value of k for which the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent.

If the lines are concurrent, they pass through same point.

Let:

$x+2y=0--eq(1)\\ 3x-4y-10=0--eq(2)\\ 5x+ky-7=0--eq(3)$

First solving equation 1 and 2 to find values of x and y

From eq(1) we find value of x and put it in eq(2)

$From \ eq(1) x+2y=0\\x=-2y\\Put x=-2y \ in \ eq(2)\\3x-4y-10=0\\3(-2y)-4y-10=0 \\-6y-4y=10\\-10y=10\\y=\frac{10}{-10}\\y=-1$

After solving we get value of y=-1

Now putting in eq(1) to get value of x

$x+2y=0\\x+2(-1)=0\\x-2=0\\x=2$

So, Value of x= 2

Now put value of x=2 and y=-1 into eq(3) to find value of k

$5x+ky-7=0\\5(2)+k(-1)-7=0\\10-k-7=0\\-k+3=0\\-k=-3\\k=3$

So, After solve the equations we get value of k=3

4 0

2 years ago