Answer:
15.5 ft
Step-by-step explanation:
The geometry of the problem can be modeled by a right triangle with hypotenuse 16 ft and one side length of 4 ft. If x represents the height of the ladder on the building, then the Pythagorean theorem tells us ...
x^2 + (4 ft)^2 = (16 ft)^2
x^2 = 240 ft^2 . . . . . . subtract 16 ft^2
x ≈ 15.5 ft . . . . . . . . . . take the square root
The top of the ladder is about 15.5 ft above the ground.
Answer:
A. 50°
Step-by-step explanation:
The external angle ACB created by tangents CA and CB is the supplement of arc AB it intercepts.
∠ACB = 180° -AB
∠ACB = 180° -130°
∠ACB = 50°
I=prt
i=(45000)(0.08)(1)
i=3600
depreciation= $3600/y
Answer:
3y= -2x-6
Step-by-step explanation:
The two points on the line are (-3,0) and (0,-2)
so you first get the gradient;
gradient= <u>change in y</u>
change in x
= <u>-2-0</u>
0-(-3)
=<u> -2</u>
3
so the answer above is the gradient
then pick one point that you used to get the gradient with, so as for me I'll pick (-3,0) and then a general point which is always (x, y)
since you have the gradient you can easily get the equation by doing this
<u>y-0</u><u>. </u><u> </u> = <u>-2</u>
x-(-3). 3
then crossmultiply to get the equation of the line
3y= -2(x+3)
3y= -2x-6
