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3 years ago

Use the graph to estimate the height of a jugglers ball that has been in the air for 1.5 seconds

Mathematics

1 answer:

strojnjashka [21]3 years ago

5 0

Answer:

The height of a jugglers ball that has been in the air for 1.5 seconds is 6.5 feet

Step-by-step explanation:

we are given height equation as

$h=-16t^2+25t+5$

where

h is the height in feet of a Juggler's ball after t seconds

now, we can plug t=1.5

$h=-16(1.5)^2+25(1.5)+5$

we can solve for h

$h=42.5-16\cdot \:1.5^2$

$h=42.5-36$

$h=6.5$

So,

The height of a jugglers ball that has been in the air for 1.5 seconds is 6.5 feet

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2 years ago

Rewrite the following quadratic functions in intercept or factored form. Show your work.

Mariulka [41]

Answer:

y = 5x (x+2)

Step-by-step explanation:

We are given the following the quadratic function and we are to rewrite it in intercept or factored form:

$y = 5x^2 + 10x$

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3 years ago

Read 2 more answers

Sketch the graph of each line. 2x-5y=25

ki77a [65]

Well, you transform the standard form equation into slope-intercept form (y=mx+b), since you can only graph it using this form.

First of all, the layout of the equation is $Ax+By=C$

So, in this case, A is 2, B is 5, and C is 25.

Now, to put it into slope intercept form, you either use the formula to change it and solve it:

$y= \frac{-A}{B}x+ \frac{C}{B} \\ y= \frac{-2}{-5}x+ \frac{25}{-5} \\ \\ y= \frac{2}{5}x-5$

Or you can just solve for y:

$2x-5y=25 \\ 2x-5y-2x=25-2x \\ -5y=-2x+25 \\ \frac{-5y}{-5}= \frac{-2x+25}{-5} \\ \\ y= \frac{2}{5}x-5$

In addition, I attached what the graph looks like.

Hope this helps!