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Nat2105 [25]
2 years ago
14

Lisa spends part of her year as a member of a gym. She then finds a better deal at another gym

Mathematics
1 answer:
Tema [17]2 years ago
5 0
I’m having trouble understanding this question.?
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Tank A has a capacity of 9.5 gallons. 6 1\3 gallons of the tank"s water are poured out. How many gallons of water are left in th
aleksandr82 [10.1K]

Answer:  Since 1/3 can't be turned into a decimal, change .5 into 1/2. In order to subtract, the fractions need a common denominator. The least common denominator is 6. So 1/3 turns into 2/6, and 1/2 turns into 3/6. Now you need to take 9 3/6 minus 6 2/6. The answer is 3 1/6 gallons of water.


Step-by-step explanation:


4 0
2 years ago
At a museum 100 posters are displayed in each of four rooms altogether how many posters are displayed
aleksandr82 [10.1K]
25 posters in each room.
5 0
3 years ago
Which is least 1.3 1 1/3 or 1.34
Evgesh-ka [11]
1.3 is the least because ur supposed to add a zero so they can all be the same place value
5 0
3 years ago
Nick had 3 1/4 bottles of water for his wrestling practice when he finished he had 124 bottles of water left he said he used to
Ostrovityanka [42]
Nick had 3 1/4 bottles before, and he had 1 2/4 bottles left, so that means
3 1/4 - 1 2/4 =  1.75
1.75= 1 3/4.
So, I do not agree with nick.

Hope this helps!!!   :)
8 0
3 years ago
A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h(t) = -16t2 + 6
katrin2010 [14]

Answer:

Part a) The height of the ball after 3 seconds is 49\ ft

Part b) The maximum height is 66 ft

Part c) The ball hit the ground for t=4 sec

Part d) The domain of the function that makes sense is the interval

[0,4]

Step-by-step explanation:

we have

h(t)=-16t^{2} +63t+4

Part a) What is the height of the ball after 3 seconds?

For t=3 sec

Substitute in the function and solve for h

h(3)=-16(3)^{2} +63(3)+4=49\ ft

Part b) What is the maximum height of the ball? Round to the nearest foot.

we know that

The maximum height of the ball is the vertex of the quadratic equation

so

Convert the function into a vertex form

h(t)=-16t^{2} +63t+4

Group terms that contain the same variable, and move the constant to the opposite side of the equation

h(t)-4=-16t^{2} +63t

Factor the leading coefficient

h(t)-4=-16(t^{2} -(63/16)t)

Complete the square. Remember to balance the equation by adding the same constants to each side

h(t)-4-16(63/32)^{2}=-16(t^{2} -(63/16)t+(63/32)^{2})

h(t)-(67,600/1,024)=-16(t^{2} -(63/16)t+(63/32)^{2})

Rewrite as perfect squares

h(t)-(67,600/1,024)=-16(t-(63/32))^{2}

h(t)=-16(t-(63/32))^{2}+(67,600/1,024)

the vertex is the point (1.97,66.02)

therefore

The maximum height is 66 ft

Part c) When will the ball hit the ground?

we know that

The ball hit the ground when h(t)=0 (the x-intercepts of the function)

so

h(t)=-16t^{2} +63t+4

For h(t)=0

0=-16t^{2} +63t+4

using a graphing tool

The solution is t=4 sec

see the attached figure

Part d) What domain makes sense for the function?

The domain of the function that makes sense is the interval

[0,4]

All real numbers greater than or equal to 0 seconds and less than or equal to 4 seconds

Remember that the time can not be a negative number

6 0
3 years ago
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