1. 1/4(x+3)=12=X+3-12*4/4=X-45/4. Then, x-45/4=0. So, 1. X=45. 2. x+4/3=12=3x-32/3*3=0*3. 3x-32=0. You add 32 to each sides. 3x=32. x=32/3. Therefore, 1. x=45 and 2. x=32/3
The answer would be 0.90 and 0.900
Answer:
The slope of the line is 6
Step-by-step explanation:
The slope of the line is given by
m=(y2-y1)/(x2-x1)
m =(8-2)/(2-1)
= 6/1
The slope of the line is 6
I got
t = 6.614
-16t^2 +1700 = 1000
-16t^2 = -700
t^2 = 43.75
You need to divide by 16 first, before you take the square root.
Domain = what values of time would be used here?
t > 0, since negative time not reasonable. And you probably need to find when the ball hits the ground, since after that time, the ball would be underground.
So the a reasonable domain is from 0 to 10.3078
Range: what values of height seem reasonable?
h > 0, because negative height would mean that the ball was underground.
And ... since the ball was "dropped" from a height of 1700 (vs being "thrown up" from a height) the maximum height would be 1700.
So ... a reasonable range would be from 0 to 1700.
Hope this helps... :)
Answer:
m∠Y = 50°
Step-by-step explanation:
let 'm' = m∠Y
m∠S + m∠Y = 90 (complementary)
5m - 210 + m = 90
6m = 300
m = 50
