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castortr0y [4]
3 years ago
8

Solve for v 160=11-v

Mathematics
1 answer:
user100 [1]3 years ago
6 0
V + 160 = 11
v = -149

Hope this helps
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What is the ratio of 1/6:6 = what
frutty [35]
1/6=16.666666...

So it is 16.6666.... to 6(or 100)

5 0
3 years ago
-6(2r +8)=-10(r-3)<br> I need the work to
barxatty [35]

Answer:

r = -39

Step-by-step explanation:

So we are trying to solve the equation for r.

-6(2r + 8) = -10(r - 3)

Divide both sides by -2. (This will make distributing much easier.)

3(2r + 8) = 5(r - 3)

Distribute the 3 on the left side and the 5 on the right side,

6r + 24 = 5r - 15

Subtract 24 from both sides.

6r = 5r - 39

Subtract 5r from both sides.

r = -39

So now we have solve the equation.

I hope you find my answer and explanation to be helpful. Happy studying. :)

5 0
3 years ago
You are given a problem that involves multiplying a positive number and a negative number.
kenny6666 [7]

Answer:

-3 and -2

Step-by-step explanation:

A better explanation...

(positive)(positive)= positive

(positive)(negative)= negative

(negative)(negative)= positive

(negative)(positive)= negative

3 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!! WILL AWARD BRAINLIEST!! THANKS!! A restaurant offers 6 choices of appetizer, 8 choices of main meal and 5 ch
gayaneshka [121]

Answer:

377 meals

Step-by-step explanation:

If you choose to have all three courses, then there are 6 choices for the first course, 8 for the second, and 5 for the last, making a total of 6*8*5=240 different possible meals.

If you choose two courses, then there are 3 options. You can pick appetizer and main meal, which would give you 6*8=48 possibilities. You can pick main meal and dessert, which would give you 8*5=40 possibilities. Finally you can pick appetizer and desert, which would give you 6*5=30 possibilities. In total these are 118 different possible meals with two courses.

Finally, you could choose 1 course, which would give you 6+8+5=19 different meals.

In total, this is 240+118+19=377 meals

8 0
3 years ago
Solve the compound inequality,
Feliz [49]

Answer:

\large\boxed{y\in(-6,\ \infty)}

Step-by-step explanation:

(1)\\2y-2>-14\qquad\text{add 2 to both sides}\\\\2y-2+2>-14+2\\\\2y>-12\qquad\text{divide both sides by 2}\\\\\dfrac{2y}{2}>\dfrac{-12}{2}\\\\y>-6

(2)\\\\4y-4\geq12\qquad\text{add 4 to both sides}\\\\4y-4+4\geq12+4\\\\4y\geq16\qquad\text{divide both sides by 4}\\\\\dfrac{4y}{4}\geq\dfrac{16}{4}\\\\y\geq4

\text{From (1) and (2) we have:}\\\\y>-6\ or\ y\geq4\to y\in(-6,\ \infty)\ \cup\ [4,\ \infty)\Rightarrow y\in(-6,\ \infty)

7 0
3 years ago
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