+0

easy question but can't seem to be getting the full marks

+1
299
1
+557

https://vle.mathswatch.co.uk/images/questions/question15619.png

Feb 13, 2018

#1
+7598
+3

Since they form a straight line....

(x + 25)° + (2x + 50)°   =   180°      Now we can find the value of  x .

Simplify the left side of the equation.

x + 25 + 2x + 50   =   180

3x + 75   =   180

Subtract  75  from both sides of the equation.

3x   =   105

Divide both sides by  3 .

x   =   35

In order for  AC  to be parallel to  EF,   2x + 50   must be equal to  5x - 55 .

2x + 50   \(\stackrel{?}{=}\)   5x - 55

We know that  x = 35 , so plug in  35  for  x .

2(35) + 50   \(\stackrel{?}{=}\)   5(35) - 55

70 + 50   \(\stackrel{?}{=}\)   175 - 55

120   \(\stackrel{?}{=}\)   120       Yes, so  AC  is parallel to  EF .

Feb 13, 2018

#1
+7598
+3

Since they form a straight line....

(x + 25)° + (2x + 50)°   =   180°      Now we can find the value of  x .

Simplify the left side of the equation.

x + 25 + 2x + 50   =   180

3x + 75   =   180

Subtract  75  from both sides of the equation.

3x   =   105

Divide both sides by  3 .

x   =   35

In order for  AC  to be parallel to  EF,   2x + 50   must be equal to  5x - 55 .

2x + 50   \(\stackrel{?}{=}\)   5x - 55

We know that  x = 35 , so plug in  35  for  x .

2(35) + 50   \(\stackrel{?}{=}\)   5(35) - 55

70 + 50   \(\stackrel{?}{=}\)   175 - 55

120   \(\stackrel{?}{=}\)   120       Yes, so  AC  is parallel to  EF .

hectictar Feb 13, 2018