Answer:
6
Step-by-step explanation:
so this is (base times hight)
(bH)
42=7x
you didvide 7 from both sides
42 divided by 7 is 6
so basically its 6
because that would be your base thats what you are missing
then its just baseically (BH) (7 times 6)= 42
if this is wrong im so sorry I just learned this good luck
Answer:
See below.
Step-by-step explanation:
The Slope of PQ is (v - z) / (w - x).
The slope of P'Q' =
(v + b) - (z + b)
--------------------- = (v - z) / (w - x)
(w + a - (x + a)
Both lines have a slope that is (v - z)/ (w - x).
So both lines are parallel.
Answer: 
<u>Step-by-step explanation:</u>
![\text{Use the distance formula: }d_AB=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\where\ (X_A, y_A)=(-3, -2)\\and\ (x_B,y_B)=(4, -7)\\\\\\d_AB=\sqrt{(-3-4)^2+[-2-(-7)]^2}\\\\.\quad =\sqrt{(-7)^2+(5)^2}\\\\.\quad =\sqrt{49+25}\\\\.\quad =\boxed{\sqrt{74}}](https://tex.z-dn.net/?f=%5Ctext%7BUse%20the%20distance%20formula%3A%20%7Dd_AB%3D%5Csqrt%7B%28x_A-x_B%29%5E2%2B%28y_A-y_B%29%5E2%7D%5C%5Cwhere%5C%20%28X_A%2C%20y_A%29%3D%28-3%2C%20-2%29%5C%5Cand%5C%20%28x_B%2Cy_B%29%3D%284%2C%20-7%29%5C%5C%5C%5C%5C%5Cd_AB%3D%5Csqrt%7B%28-3-4%29%5E2%2B%5B-2-%28-7%29%5D%5E2%7D%5C%5C%5C%5C.%5Cquad%20%3D%5Csqrt%7B%28-7%29%5E2%2B%285%29%5E2%7D%5C%5C%5C%5C.%5Cquad%20%3D%5Csqrt%7B49%2B25%7D%5C%5C%5C%5C.%5Cquad%20%3D%5Cboxed%7B%5Csqrt%7B74%7D%7D)
For this one, an easy way to solve it is to break it down into parts. First lets convert 5 feet to inches.
Multiply the number of feet by how many inches are in one foot.
5 times 12= 60 in
Now just add the like terms
60 in + 2 in= 62 inches.
Now we see that Sarah is 62 inches tall, now we can convert to centimeters the same way.
62 times 2.54= 157.48
So that makes C your answer, Sarah is 157.48 cm tall.
Answer:
He is 5.6 km away
Step-by-step explanation:
he starts traveling away from base and is 8.3 km away to start with
then he turns west and travels 13.9 km away
what this mean is that you have to subtract the distance he turned from the distance he started off with
ex: 13.9 - 8.3 = 5.9
to get your answer of how far away he is from the base
