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3 years ago

R(x)=x^3+5x^2-2x-24 Plot the zero(s) of the function.

Mathematics

1 answer:

nevsk [136]3 years ago

6 0

:) Brainliest pls?

Answer:

The zeros are {2, -3, -4} which need to be plotted on the x-axis.

Step-by-step explanation:

I'll find the zeros, aka x-intercepts, and you could probably graph them.

To find the zeros, let's factor this polynomial:

r(x) = (x - 2)(x^2+7x+12)

r(x) = (x - 2)(x + 3)(x + 4)

The zeros are {2, -3, -4} which need to be plotted on the x-axis.

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Yuri [45]

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3 years ago

Read 2 more answers

I need help, an answer or how to- will be fine, thank you. (i’ll mark brainliest if i can)

siniylev [52]

by solving the first equation you'll get x= -24/13

if you solve the second option (B) then you'll the same result i.e x=-24/13

so option C is correct

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2 years ago

Solve for 'p'

Darina [25.2K]

$\left(\dfrac{9a^{-4}}{25}\right)^p=\dfrac3{5a^2}$
$\left(\dfrac9{25a^4}\right)^p=\dfrac3{5a^2}$

One way to find $p$ is to notice that the part on the left hand side within the parentheses can be written as a square:

$\dfrac9{25a^4}=\dfrac{3^2}{5^2(a^2)^2}=\left(\dfrac3{5a^2}\right)^2$

So you could write

$\left(\left(\dfrac3{5a^2}\right)^2\right)^p=\left(\dfrac3{5a^2}\right)^{2p}=\dfrac3{5a^2}$

Since these two expressions are equal, and the bases are the same, you know the exponents must be equal, with the exponent on the right hand side being 1. So $2p=1$ , or $p=\dfrac12$ .

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3 years ago

I need help with this problem please help me find the answer

olga55 [171]

It’s 14.

The five numbers will be: 12, 13, 14, 15, and 17 so 14 is the median.

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3 years ago

Hat is the length of the hypotenuse, x, is a Pythagorean triple? 22 29 41 42

aniked [119]

Of your numbers listed, 29 and 41 are the lengths of the hypotenuse of a Pythagorean triple.

Those triples are
(20, 21, 29)
(9, 40, 41)

_____
The On-Line Encyclopedia of Integer Sequences (OEIS) lists the hypotenuses of primitive triples as sequence number A020882.

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3 years ago