Answer:
Step-by-step explanation:
i think that the equations might be y = x - 7
Answer:
r = {-8, -4}
Step-by-step explanation:
Simplifying
r2 = -32 + -12r
Solving
r2 = -32 + -12r
Solving for variable 'r'.
Reorder the terms:
32 + 12r + r2 = -32 + -12r + 32 + 12r
Reorder the terms:
32 + 12r + r2 = -32 + 32 + -12r + 12r
Combine like terms: -32 + 32 = 0
32 + 12r + r2 = 0 + -12r + 12r
32 + 12r + r2 = -12r + 12r
Combine like terms: -12r + 12r = 0
32 + 12r + r2 = 0
Factor a trinomial.
(8 + r)(4 + r) = 0
Subproblem 1
Set the factor '(8 + r)' equal to zero and attempt to solve:
Simplifying
8 + r = 0
Solving
8 + r = 0
Move all terms containing r to the left, all other terms to the right.
Add '-8' to each side of the equation.
8 + -8 + r = 0 + -8
Combine like terms: 8 + -8 = 0
0 + r = 0 + -8
r = 0 + -8
Combine like terms: 0 + -8 = -8
r = -8
Simplifying
r = -8
Subproblem 2
Set the factor '(4 + r)' equal to zero and attempt to solve:
Simplifying
4 + r = 0
Solving
4 + r = 0
Move all terms containing r to the left, all other terms to the right.
Add '-4' to each side of the equation.
4 + -4 + r = 0 + -4
Combine like terms: 4 + -4 = 0
0 + r = 0 + -4
r = 0 + -4
Combine like terms: 0 + -4 = -4
r = -4
Simplifying
r = -4
Solution
r = {-8, -4}
Height of the woman = 5 ft
Rate at which the woman is walking = 7.5 ft/sec
Let us assume the length of the shadow = s
Le us assume the <span>distance of the woman's feet from the base of the streetlight = x
</span>Then
s/5 = (s + x)/12
12s = 5s + 5x
7s = 5x
s = (5/7)x
Now let us differentiate with respect to t
ds/dt = (5/7)(dx/dt)
We already know that dx/dt = 7/2 ft/sec
Then
ds/dt = (5/7) * (7/2)
= (5/2)
= 2.5 ft/sec
From the above deduction, it can be easily concluded that the rate at which the tip of her shadow is moving is 2.5 ft/sec.
Isolate the x
divide 3 from both sides
3x/3 > 6/3
x > 6/3
x > 2 is your answer
Note: because you are not dividing by a negative number, you do not need to flip the sign
hope this helps
You can use calculus ( related rates). Given the rate os change of the radius for example you can find the rate of change of volume using differential calculus.
