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Mnenie [13.5K]
3 years ago
14

What are the zeros of f(x)=x^2+3x-10

Mathematics
2 answers:
Lelu [443]3 years ago
7 0

Answer:

x= -5 and x= 2

Step-by-step explanation:

vekshin13 years ago
4 0

You must remember that a polynomial is written like so...

ax^2 + bx + c

In this case...

a = 1

b = 3

c = -10

To factor you must find two numbers who both add up to b (3) AND multiply to c (-10)

-2 + 5 = 3

-2 * 5 = -10

so...

(x - 2)(x + 5)

To find the zero you must set each factor equal to zero and solve for for x like so...

x - 2 = 0

x = 2

x + 5 = 0

x = -5

Hope this helped!

~Just a girl in love with Shawn Mendes

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5 would be the answer to it
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Step-by-step explanation:

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3 years ago
The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped
kakasveta [241]

Check the picture below.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2  = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

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10/18 because 18-8=10 and the total amount is 18 so the fraction is 10/18
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