Please help??!?!??? Bplease
1 answer:
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Answer: a = 4, b = 163,422
<u>Step-by-step explanation:</u>
P is the new population, P₀ is the initial population, k is the growth rate, t is the time elapsed.
Part a) What do we know?
P = 7003
P₀ = unknown
k = .21 <em>(converted 21% into a decimal)</em>
t = 36 yrs <em>(2006 - 1970) </em>
<em>need to solve for P₀</em>
7003 = P₀ (1920)
3.6 = P₀
rounded to the nearest whole number = 4
Part b) What do we know?
P = unknown
P₀ = 7003
k = .21 <em>(converted 21% into a decimal)</em>
t = 15 yrs <em>(2021 - 2006) </em>
<em>need to solve for P₀</em>
P = 7003(23.33)
P = 163,422.46
rounded to the nearest whole number = 163,422
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Answer: 2, -1
<u>Step-by-step explanation:</u>
eˣ² = eˣ * e²
eˣ² = eˣ⁺²
x² = x + 2
x² - x - 2 = 0
(x - 2)(x + 1) = 0
x - 2 = 0 or x + 1 = 0
x = 2 or x = -1
Check both answer for validity:
eˣ² = eˣ⁺²
e⁴ = e⁴ or e¹ = e¹
TRUE TRUE
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Answer: 0 = ln 1
<u>Step-by-step explanation:</u>
-10 = ln 1 - 10 <em>log rules say division is subtraction</em>
0 = ln 1
Answer:
18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144
b. The maximum height = 400 feet
c. Attached graph
19) The rocket will reach the maximum height after 4 seconds
20) The rocket hits the ground after 9 seconds
Step-by-step explanation:
* Lets study the rule of motion for an object with constant acceleration
# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time
and a is the acceleration of gravity
# The vertical distances h in x second is h(x) - h(0), where h(0)
is the initial height of the object above the ground
∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial
velocity, a is the acceleration of gravity (32 feet/second²) and x
is the time
18)a.
∵ The value of a = -32 ft/sec² ⇒ negative because the direction
of the motion
is upward
∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16
∴ h(x) = vx - 16x² + h(0)
∴ h(x) = -16x² + vx + h(0) ⇒ proved
* Find the height of the object after x seconds from the ground
∵ h(0) = 144 and v = 128 ft/sec
∴ h(x) = -16x² + 128x + 144
b.
* At the maximum height h'(x) = 0
∵ h'(x) = -32x + 128
∴ -32x + 128 = 0 ⇒ subtract 128 from both sides
∴ -32x = -128 ⇒ ÷ -32
∴ x = 4 seconds
- The time for the maximum height = 4 seconds
- Substitute this value of x in the equation of h(x)
∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet
c. Attached graph
19)
- The object will reach the maximum height after 4 seconds
20)
- When the rocket hits the ground h(x) = 0
∵ h(x) = -16x² + 128x + 144
∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16
∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x
∵ x² - 8x - 9 = 0
∴ (x - 9)( x + 1) = 0
∴ x - 9 = 0 OR x + 1 = 0
∴ x = 9 OR x = -1
- We will rejected -1 because there is no -ve value for the time
* The time for the object to hit the ground is 9 seconds
