Add the prices together
Then divide by 3.
*6.25 + 3.25 = 9.50/3 = 2 of them pay 3.16 while the other pays 3.17.
Answer:
The ladder will reach 4.6 ft up the tree.
Step-by-step explanation:
When we draw out our triangle, we should see that we are given the horizontal leg and are trying to find the vertical leg. We use tan∅ to help solve:
tan53° = x/3.5
3.5tan53° = x
x = 4.64466
IF IT EQUALS 0
1. find two number that multiplied to 48 and adds to 14, which are 6 and 8.
2. substitute the new numbers in with x to get x^2 + 6x + 8x + 48.
3. factor out the x and the 8 to get x(x+6)+8(x+6).
4. x = -6, x = -8
IF IT DOES NOT EQUAL 0
then (x+6)*(x+8) is your answer.
(a) When f is increasing the derivative of f is positive.
f'(x) = 15x^4 - 15x^2 > 0
15x^2(x^2 - 1)> 0
x^2 - 1 > 0 (The inequality doesn't flip sign since x^2 is positive)
x^2 > 1
Then f is increasing when x < -1 and x > 1.
(b) The f is concave upward when f''(x) > 0.
f''(x) = 60x^3 - 30x > 0
30x(2x^2 - 1) > 0
x(2x^2 - 1) > 0
x(x^2 - 1/2) > 0
x(x - 1/sqrt(2))(x + 1/sqrt(2)) > 0
There are four regions here. We will check if f''(x) > 0.
x < -1/sqrt(2): f''(-1) = -30 < 0
-1/sqrt(2) < x < 0: f''(-0.5) = 7.5 > 0
0 < x < 1/sqrt(2): f''(0.5) = -7.5 < 0
x > 1/sqrt(2): f''(1) = 30 > 0
Thus, f''(x) > 0 at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
Therefore, f is concave upward at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
(c) The horizontal tangents of f are at the points where f'(x) = 0
15x^2(x^2 - 1) = 0
x^2 = 1
x = -1 or x = 1
f(-1) = 3(-1)^5 - 5(-1)^3 + 2 = 4
f(1) = 3(1)^5 - 5(1)^3 + 2 = 0
Therefore, the tangent lines are y = 4 and y = 0.
The common ratio of this geometric sequence is D.9
