Answer:
C. 132
Step-by-step explanation:
We assume all angle measures are in degrees. The vertical angles are equal in measure, so ...
4x = x +36
3x = 36
x = 12
x +36 = 48
y = 180 -48 = 132
The value of y is 132.
Answer:
same amount at 2 months. he'd pay 140.
Step-by-step explanation:
35(month) +35(month) =70(joining fee). 70+70=140. that's two months at the first gym.
the second gym is also 140 for the two months because each month is $70. and 70+70=140
A -76x+76=-76x+76−76x+76=−76x+76minus, 76, x, plus, 76, equals, minus, 76, x, plus, 76 (Choice B) B 76x+76=-76x+7676x+76=−76x+76
Pie
Answer:
b
Step-by-step explanation:
i could be wrong
Answer:
Ms. Thomas was driving at constant rate of 52 miles/hour.
Step-by-step explanation:
Given:
Total time to travel (t) = 45 minutes
Distance drove (d) = 39 miles
we need to find the constant rate in miles per hour at which she was driving.
Solution:
Now we know that;
We need to find constant rate at miles per hour;
But time is given in minutes.
So we will convert minutes into hour by dividing by 60 we get;
time 
Now we know that;
Distance is equal to rate times time.
framing in equation form we get;
distance 
Or
rate 
Hence Ms. Thomas was driving at constant rate of 52 miles/hour.
Answer:
y = x + 7
Step-by-step explanation:
The slope-intercept form of a line is:
y = mx + b
where m is the slope and b is the y-intercept.
Looking at the graph, we can see that the line intersects the y-axis at y = 7. So 7 would be our y-intercept.
To find the slope, we would divide the rise of the line by the run. Or m = rise/run. From looking at the graph, we can see that for every 1 unit the line moves in the x-direction, the line moves in the y-direction by 1 unit. Therefore, the rise would be 1 and the run would be 1. 1/1 = 1 so the slope of the line would be 1.
Plugging in 7 for b and 1 for m into the equation for the slope-intercept form, we get:
y = x + 7
So that would be the equation for the line in slope-intercept form.
I hope you find my answer and explanation to be helpful. Happy studying.
