Answer:
yes, ±2
Step-by-step explanation:
The x-intercepts are found by setting y=0 and solving for x:
x^2/4 = 1
x^2 = 4
x = ±√4
x = ±2
The x-values of interest are -2 and +2.
Assume ladder length is 14 ft and that the top end of the ladder is 5 feet above the ground.
Find the distance the bottom of the ladder is from the base of the wall.
Picture a right triangle with hypotenuse 14 feet and that the side opposite the angle is h. Then sin theta = h / 14, or theta = arcsin 5/14. theta is
0.365 radian. Then the dist. of the bot. of the lad. from the base of the wall is
14cos theta = 14cos 0.365 rad = 13.08 feet. This does not seem reasonable; the ladder would fall if it were already that close to the ground.
Ensure that y ou have copied this problem accurately from the original.
Answer:
X=10.5/F-2.5
Step-by-step explanation:
A, C, and E.
you can check them by putting into y=mx+b form. if m is negative then as domain increases range decreases.
1. 14.3-2*5²÷5=14.3-10=4.3
2. Result is Twenty one
