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nikitadnepr [17]
2 years ago
14

Round the answer to the nearest cent. 5 1/2% tax on $20 is $

Mathematics
1 answer:
dybincka [34]2 years ago
3 0

Answer:

goodmorning b

Step-by-step explanation:

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Answer:

150% of 2/3 km

150/100 * 2/3

50/50

= 1 km

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The system of equations has 2x-y=2(x+1) y=-2
sleet_krkn [62]

Answer:

A.

Step-by-step explanation:

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Consider the function represented by the equation x - y = 3. What is the equation written in function notation, with x
Svet_ta [14]

Independent Variable Y-3(x)

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2 years ago
The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time
cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

1\leq t\geq \sqrt{2}

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

h =20-5t^{2}-----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height h = 15m  

Put height value in equation 1

15 =20-5t^{2}

5t^{2} =20-15

5t^{2} =5

t^{2} =1

t =\pm1

We know that the time is always positive, therefore t=1

when the stuntman is 10m above the ground.

height h = 10m  

Put height value in equation 1

10 =20-5t^{2}

5t^{2} =20-10

5t^{2} =10

t^{2} =\frac{10}{5}

t^{2} =2

t=\pm\sqrt{2}

t=\sqrt{2}

Therefore ,time interval of camera film him is 1\leq t\geq \sqrt{2}

7 0
3 years ago
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