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ehidna [41]
3 years ago
14

Eliminate the parameter and write a rectangular equation for.

Mathematics
1 answer:
Artyom0805 [142]3 years ago
8 0
Eliminate the parameter t:t=y/2
x=y^2/2+3
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What is the height of an object thrown when time is 6 seconds? Also explain how u got your answer
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Answer:

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Step-by-step explanation:

Let

h(t) ---> the height of an object in meters

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we have

h(t)=-2.6t^2+40t+800

For t=6 sec

substitute in the quadratic equation  the value of t and solve for h

h(6)=-2.6(6)^2+40(6)+800

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Step-by-step explanation:

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3 years ago
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\dfrac{27}{54}=\dfrac{27:9}{54:9}=\dfrac{3}{6}=\dfrac{3:3}{6:3}=\dfrac{1}{2}

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Step-by-step explanation:

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