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boyakko [2]
3 years ago
13

Solve for x. 2/3(x - 7) = -2

Mathematics
2 answers:
lesya [120]3 years ago
6 0
First you distribute 2/3 into parenthesis, then simplify the equation by multiplying both sides by the common denominator 3. then reduce with 3,3, then multiply. move constant to the right side change its sign. Then find the sum x = 4
son4ous [18]3 years ago
4 0
2/3x -14/3 = -2  -foiled

2/3x = -6/3 +14/3 -common denominator

2/3x = 8/3

x = 8/3 * 3/2 

x = 24/6 

x= 4

You might be interested in
3w + 2 = 7w Please help
mash [69]
3w +2 =7w
-3w       -3w
2=4w
Divide each side by 4
1/2=w

Check your answer:
3 (1/2) + 2 = 7 (1/2)
(3/2) + (4/2) = (7/2)
(7/2) = (7/2)

Hope this helps.
8 0
3 years ago
Read 2 more answers
Some one help me please
levacccp [35]
I think the answer is 6t=9
7 0
2 years ago
Which ordered pairs are in the solution set of the system of linear inequalities? y > x y < x + 1 (5, –2), (3, 1), (–4, 2)
steposvetlana [31]

To check which ordered pair (point) is in the solution set of the system of given linear inequalities y>x, y<x+1; we just need to plug given points into both inequalities and check if that point satisfies both inequalities or not. If any point satisfies both inequalities then that point will be in solution.

I will show you calculation for (5,-2)

plug into y>x

-2>5

which is clearly false.

plug into y<x+1

-2<5+1

or -2<6

which is also false.

hence (5,-2) is not in the solution.

Same way if you test all the given points then you will find that none of the given points are satisfying both inequalities.

Hence answer will be "No Solution from given choices".

5 0
3 years ago
Read 2 more answers
Drag the tiles to list the sides of △MNO from shortest to longest.
sweet [91]

The smaller the angle subtended by a side, the smaller the length of the

side.

The correct responses are;

Question 1: The list of sides from shortest to longest are;

  • MO/Shortest MO/Medium and MO/Longest

a) <u>Friday</u>

b) <u>70 minutes</u>

c) <u>40%</u>

d) Yes<u>,</u> <u>the sum of the </u><u>mean</u><u> number of </u><u>minutes spent</u><u> on </u><u>aerobic</u><u> training and the mean number of minutes spent on </u><u>strength</u><u> training is equal to the mean </u><u>total</u><u> number of minutes spent </u><u>training.</u>

From the given diagram, we have, the measure of the third angle, ∠O, is

found as follows;

∠O = 180° - 54° - 61° = 65°

Therefore, ∠O = The largest angle

We get;

The longest side is opposite the largest angle, which gives;

The shortest side is the side opposite ∠N (54°)= \frac{}{MO}

The next shortest side is the side opposite ∠M(61°) = \frac{}{NO}

The longest side is the side opposite ∠O(65°) = \frac{}{MN}

a) The time spent training on Tuesday = 60 + 10 = 70 minutes

The time spent training on Thursday = 50 + 30 = 80 minutes

The time spent training on Friday = 45 + 40 = 85 minutes

Therefore, the day the athlete spent the longest total amount of time training is on <u>Friday</u>

b) The time spent training on Monday = 10 + 20 = 30 minutes

The time spent training on Wednesday = 20 + 15 = 35 minutes

Therefore, we get;

30, 35, 70, 80, and 85

The median total number of minutes the athlete spent training each day = <u>70 minutes</u>

<u />

c) The time spent strength training = 20 + 10 + 15 + 30 + 45 = 120

The total number of minutes the athlete spent training = 70 + 80 + 85 + 30 + 35 = 300

The  percentage spent on strength training = \frac{120}{300} × 100 = \frac{40}%

d) The mean number of minutes spent on strength training is found as follows;

Mean_{strength} =\frac{120}{5} =24

The mean number of minutes spent on aerobic training is found as follows;

Mean_{aerobic} =\frac{10+60+20+50+40}{5} =36

Mean_{strength} +Mean_{aerobic} =24+36=60

The mean total number of minutes spent training, Mean_{total} = \frac{300}{5} = 60

Therefore;

  • Mean_{strength}+Mean_{aerobic} = Mean_{total} \\

Learn more here:

brainly.com/question/2962546

4 0
2 years ago
What is the greatest common factor of: 6c^3d - 12c^2d^2 + 3cd?
VMariaS [17]
I hope this helps you

4 0
3 years ago
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