I need help on this math problem.

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

7 0

110 - 4x = x + 10
100 = 5x
x = 20

2x - 20 = 3y - 10
Plug in x
2(20) - 20 = 3y - 10
40 - 20 = 3y - 10
20 = 3y - 10
30 = 3y
y = 10

GREYUIT [131]3 years ago

5 0

SOLUTION:

First we will find ( x ) so that we can then find ( y ).

To find x -

Opposite sides of a paralellogram are congruent ( equivalent to each other ).

( x + 10 ) and ( 110 - 4x ) are opposite sides.

Therefore:

x + 10 = 110 - 4x

x + 4x = 110 - 10

5x = 100

x = 100 / 5

x = 20

To find y -

From this, we will substitute the value of ( x ) from above into our equation for ( y ).

Opposite sides of a paralellogram are congruent ( equivalent to each other ).

( 2x - 20 ) and ( 3y - 10 ) are opposite sides.

Therefore:

2x - 20 = 3y - 10

3y - 10 = 2x - 20

3y - 10 = 2 ( 20 ) - 20

3y = 40 - 20 + 10

3y = 30

y = 30 / 3

y = 10

FINAL ANSWER:

Therefore, the answer is:

y = 10

Hope this helps! :)
Have a lovely day! <3

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Graph the circle (x+6)^2 + (y+5)^2 =9

Andrei [34K]

Answer:

This a circle centered at the point $(-6,-5)$ , and of radius "3" as it is shown in the attached image.

Step-by-step explanation:

Recall that the standard formula for a circle of radius "R", and centered at the point $(x_0,y_0)$ is given by:

$(x-x_0)^2+(y-y_0)^2=R^2$

Therefore, in our case, by looking at the standard equation they give us, we extract the following info:

1) $R^2 = 9\,\,\,then\,\,\,R=3$ since the radius must be a positive number and ( $R=-3$ ) is not a viable answer.

2) $x_0=-6$ for ( $x-x_0$ ) to equal $(x+6)$

3) $y_0=-5$ for ( $y-y_0$ ) to equal $(y+5)$

Therefore, we are in the presence of a circle centered at the point $(-6,-5)$ , and of radius "3". That is what we draw as seen in the attached image.