Make a table of ordered pairs for the equation. y= 1 2 x−3 PLEASE HELP !!

1 answer:

6 0

All you need to do is plug in some x values and find the correspoding y value
when x=0, y=-3
when x=1, y=9
when x=2, y=21
when x=-1, y=-15

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?/0.6=1.4<br>how to show the solution

den301095 [7]

To find the solution to this problem, you would do the opposite of division which is multiplication.

Use the terms you have to plug into your new equation;

0.6 x 1.4 = .84

To check your work you would plug in .84 where the '?' is;

.84/.6= 1.4

There you have the original equation you began with.

Therefore, .84 would be your final answer.

8 0

2 years ago

Solve with a system of equations: Two angles are supplementary if the sum of their measures are 180 degrees. If one is 90 degree

julsineya [31]

Answer:

The measures of the angles are 150° and 30°.

Step-by-step explanation:

Let x and y represent the measures of the angles, with x representing the larger angle.

x + y = 180 . . . . . . the two angles are supplementary

x = 90 + 2y . . . . . one is 90° more than twice the other

___

Substituting the expression given by the second equation into the first, we have ...

(90 +2y) +y = 180

3y = 90 . . . . . . . . . . collect terms, subtract 90

y = 30 . . . . . . . . . . . divide by the coefficient of y

x = 180 -y = 150

The measures of the angles are 150° and 30°.

6 0

3 years ago

1500 soldiers arrive at a training camp. A few soldiers desert the camp. The drill sergeants divide the remaining soldiers into

iren [92.7K]

Answer:

The number of deserters is 34.

Step-by-step explanation:

We have to calculate the number of desertors in a group of 1500 soldiers.

The sergeant divides in groups of different numbers and count the lefts over.

If he divide in groups of 5, he has on left over. The amount of soldiers grouped has to end in 5 or 0, so the total amount of soldiers has to end in 1 or 6.

If he divide in groups of 7, there are three left over. If we take 3, the number of soldiers gruoped in 7 has to end in 8 or 3. The only numbers bigger than 1400 that end in 8 or 3 and have 7 as common divider are 1428 and 1463.

If we add the 3 soldiers left over, we have 1431 and 1466 as the only possible amount of soldiers applying to the two conditions stated until now.

If he divide in groups of 11, there are three left over. We can test with the 2 numbers we stay:

$(1431-3)/11=1428/11=129.82\\\\(1466-3)/11=1463/11=133$