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Oduvanchick [21]
3 years ago
15

your weight on the moon is about 1/6 of your weight on earth write and solve an equation to show how much a person weighs on ear

th if he weighs 16 pounds on the moon.how could you check that your answer is reasonable
Mathematics
1 answer:
pav-90 [236]3 years ago
3 0
16 * 6 = x. 

If one sixth of your weight is 16, multiply it by 6. 

16 x 6 = 96.
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9. 1/3(x+6) =8
babunello [35]
9.       ¹/₃(x + 6) = 8
    ¹/₃(x) + ¹/₃(6) = 8
           ¹/₃x + 2 = 8
           <u>       - 2  - 2</u>
            3 · ¹/₃x = 6 · 3
                     x = 18

15.       ¹/₅(x + 10) = 6
      ¹/₅(x) + ¹/₅(10) = 6
                ¹/₅x + 2 = 6
                <u>       - 2  - 2</u>
                 5 · ¹/₅x = 4 · 5
                         x = 20

20.       ¹/₈(24x + 32) = 10
      ¹/₈(24x) + ¹/₈(32) = 10
                     3x + 4 = 10
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                           <u>3x</u> = <u>6</u>
                            3     3
                             x = 2

32.        5 - ¹/₂(x - 6) = 4
      5 - ¹/₂(x) - ¹/₂(-6) = 4
              5 - ¹/₂x + 3 = 4
              5 + 3 - ¹/₂x = 4
                    8 - ¹/₂x = 4
                 <u>- 8            - 8</u>
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33.      ²/₃(3x - 6) = 3
     ²/₃(3x) - ²/₃(6) = 3
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                <u>    + 4 + 4</u>
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8 0
3 years ago
Read 2 more answers
Given: angle 1 and angle 2 form a linear pair,
iris [78.8K]

Answer:

Look at the proof down

Step-by-step explanation:

The given is;

→ ∠1 and ∠2 form a linear pair

→ ∠1 ≅ ∠3

We want to prove;

→ ∠2 and ∠3 are supplementary

<em>We will write the proof in like a table</em>

1. ∠1 and ∠2 formed a linear pair  ⇒  1. Given

2. m∠1 + m∠2 = 180°  ⇒ 2. Sum of angles on a straight line

3. ∠1 and ∠2 are supplementary angles  ⇒  3. Supplementary angles add up to 180°

4. ∠1 ≅ ∠3  ⇒  4. Given

5. m∠2 + m∠3 = 180°  ⇒  5. Substitution method

6. ∠3 is a supplement of ∠2  ⇒  6. Supplement of equal angles

7. ∠2 and ∠3 are supplementary  ⇒  7. Proved

8 0
3 years ago
One watering system needs about three times as long to complete a job as another warning system when both systems operate at the
algol13

Answer:

One watering system requires 36 minutes to complete the job alone while another watering system requires 12 minutes to complete the job alone.

Step-by-step explanation:

Given:

Both the system can complete the job = 9 minutes

We need find the time required by each system to do the job.

Solution:

Let the time required by another watering system to complete the job be 'x' mins.

Now given:

One watering system needs about three times as long to complete a job as another watering system.

Time required by one watering system = 3x

Rate to complete the job by another watering system = \frac{1}{x}\ job/min

Rate to complete the job by One watering system = \frac{1}{3x}\ job/min

Rate at which both can complete the job = \frac{1}{9}\ job/min

So we can say that;

Rate at which both can complete the job is equal to sum of Rate to complete the job by another watering system and Rate to complete the job by One watering system.

framing in equation form we get;

\frac{1}{x}+\frac{1}{3x}=\frac19

Now taking LCM to make the denominator common we get;

\frac{1\times3}{x\times 3}+\frac{1\times1}{3x\times1}=\frac19\\\\\frac{3}{3x}+\frac{1}{3x}=\frac19\\\\\frac{3+1}{3x}=\frac{1}{9}\\\\\frac{4}{3x}=\frac{1}{9}

By Cross multiplication we get;

4\times9 =3x\\\\3x =36

Dividing both side by 3 we get;

\frac{3x}{3}=\frac{36}{3}\\\\x=12\ min

Time required by One watering system = 3x =3\times 12 =36\ min

Hence One watering system requires 36 minutes to complete the job alone while another watering system requires 12 minutes to complete the job alone.

6 0
3 years ago
Number 2 rewrite as a single power
Paul [167]

Answer:

-625

Step-by-step explanation:

I think so..

4 0
3 years ago
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e-lub [12.9K]
This is just another way of asking "what is the volume of a sphere of 6cm"
The volume of a sphere is V= \frac{3}{4} \pi  R^{3}
The R is 3cm in this problem since the diameter is twice the radius. For a 3cm radius sphere this is 113.1cm^3
3 0
3 years ago
Read 2 more answers
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