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melamori03 [73]
3 years ago
6

Problem: 89.012 / 22.0 How many sig figs in the final answer?

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
5 0
3 sig figs in the final answer since the denominator has 3 sig figs. So the answer would be 4.05
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A man is driving to pick his daughter from school. He waits for his daughter to be done, then he drives back home.

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1. A farmer is painting his silo. A typical can of paint covers 400 squared meters.
Maksim231197 [3]

Answer:

1. 6/10 – 2/5 =

Ans: 2/10, which can be reduced to 1/5.

2. If a tank is 5/8 filled with solution, how much of the tank is empty?

Ans: 3/8 of the tank is empty. Since the whole tank would equal 8/8, or 1, and 5/8 of it is filled, then that

means 3/8 of it remains empty.

3. 1/2 x 3/5 x2/3 =

Ans: 6/30, which can be reduced to 1/5.

4. 5/9 ÷ 4/11 =

Ans: 55/36. You cannot reduce this fraction any further.

5. Convert 27/4 to a decimal.

Ans: 6.75. This answer is arrived at by dividing 4 into 27.

6. Convert 0.45 to a fraction.

Ans: 45/100, which can be reduced to 9/20.

7. 4.27 x 1.6 =

Ans: 6.832

8. 6.5 ÷ 0.8 =

Ans: 8.125

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Step-by-step explanation:

3 0
3 years ago
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

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

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