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ycow [4]
2 years ago
5

Please Help!! I will mark brainliest for the correct answer!!

Mathematics
2 answers:
Aneli [31]2 years ago
6 0

Step-by-step explanation:

Please Help!! I will mark brainliest for the correct answer!!

A parabola has its vertex at (2, -3) and its y-intercept at (0,5)

Write an equation of the parabola in the form y= a(x-h)^2+k

marshall27 [118]2 years ago
3 0

~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill

\stackrel{vertex}{(\stackrel{h}{2}~~,~~\stackrel{k}{-3})}\qquad y=a(x-2)^2-3~\hfill \textit{we also know that} \begin{cases} x=0\\ y=5 \end{cases} \\\\\\ 5=a(0-2)^2-3\implies 5=4a-3\implies 8=4a\implies \cfrac{8}{4}=a\implies 2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=2(x-2)^2-3~\hfill

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timurjin [86]

Answer:

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Step-by-step explanation:

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1. What is the value of f(3) and what does it represent in the context of this example?
Marysya12 [62]

Answer/step-by-step explanation:

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4 years ago
If an object is projected upward with an initial velocity of 123 ft per​ sec, its height h after t seconds is h=−16t2+123t. Find
Alinara [238K]

Answer:

Height after 5 seconds is <em>215 ft</em>

Step-by-step explanation:

Given that:

The initial velocity of object which is projected upwards is 123 ft/sec.

Height of the object, <em>h</em> after time <em>t </em> in seconds,  is given as:

h=-16t^2+123t

Here, <em>t </em>will always be positive, so 123t and 16t^2 will also be positive.

But coefficient of 16t^2 is negative, that means something is subtracted from the positive term 123t.

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Height of object after 5 seconds.

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Answer:

Whole numbers less than 5 may be termed as Rational numbers.

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