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Verizon [17]
1 year ago
11

f(x) = ax^3 + bx^2 + cx + d In the polynomial above, a, b, c, and d are constants. If f(–5) = 3, which of the following must be

true about f(x)? A x – 3 is a factor of f(x). B The remainder when f(x) is divided by x + 5 is 3. C x + 2 is a factor of f(x). D x + 5 is a factor of f(x).
Mathematics
1 answer:
Amiraneli [1.4K]1 year ago
8 0

The remainder when f(x) is divided by x + 5 is 3.

What is an Equations?

Equations are mathematical statements with two algebraic expressions on either side of an equals (=) sign. It illustrates the equality between the expressions written on the left and right sides. To determine the value of a variable representing an unknown quantity, equations can be solved. A statement is not an equation if there is no "equal to" symbol in it. It will be regarded as an expression.

we discovered that the polynomial equation is the equation that results from equating a polynomial function to zero. A linear equation results from equating a function to zero if it is a linear polynomial. We get a quadratic equation if the function is a quadratic polynomial.

f(x) = ax^3 + bx^2 + cx + d

In the polynomial above, a, b, c, and d are constants and f(-5)=3.

- 5 is the root of the given polynomial when remainder is 3 so we get the one factor of the given polynomial and that is,

x-(-5)=(x+5)

That is if f(-5)=3 it means that the remainder when f(x) is

divided by (x-(-5))=x+5 is 3

Hence, The remainder when f(x) is divided by x + 5 is 3

Learn more about equations, by the following link

brainly.com/question/2972832

#SPJ1

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Subtract 6(x^2-xy) from 3x(x-y)
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6 0
3 years ago
________is a ratio comparing two quantities with different kinds of units
V125BC [204]

Answer:

Rate

Step-by-step explanation:

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for example:

When i compare two quantities, 1st is a distance quantity and 2nd is a time quantity i.e.

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5 0
4 years ago
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Find Exact Answers, Show your work. BRANLIEST ASAP!! THANK YOU​
4vir4ik [10]

$\cos\theta=\frac{\text{5}}{\text{13}},  \ \sin\theta=\frac{\text{12}}{13}, \ \tan\theta=\frac{\text{12}}{\text{5}}

$\sec\theta=\frac{\text{13}}{\text{5}},\ \csc \theta=\frac{\text{13}}{\text{12}} , \ \cot \theta=\frac{\text{5}}{\text{12}}

Solution:

Given triangle is a right triangle.

Base = 5, Hypotenuse = 13, Height = ?

Height (h) = opposite

<em>In right triangle, square of the hypotenuse is equal to the sum of the squares of the other two sides.</em>

5^2+h^2=13^2

25+h^2=169

h^2=144

Taking square root on both sides.

h = 12

$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}

$\cos\theta=\frac{\text{5}}{\text{13}}

$\sin\theta=\frac{\text{Opposite}}{\text{Hypotenuse}}

$\sin\theta=\frac{\text{12}}{13}

$\tan\theta=\frac{\text{Opposite}}{\text{Base}}

$\tan\theta=\frac{\text{12}}{\text{5}}

$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base }}

$\sec\theta=\frac{\text{13}}{\text{5}}

$\csc \theta=\frac{\text{Hypotenuse}}{\text{Opposite }}

$\csc \theta=\frac{\text{13}}{\text{12}}

$\cot \theta=\frac{\text{Base }}{\text{Opposite}}

$\cot \theta=\frac{\text{5}}{\text{12}}

6 0
4 years ago
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