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LUCKY_DIMON [66]
3 years ago
10

When g(x) is divided by x + 4 the remainder is 0. Given g(x)=x^4+3x^3-6x+8 which conclusion about g(x) is true?

Mathematics
1 answer:
Leona [35]3 years ago
5 0
Check your textbook on the Remainder Theorem.

any divisor that yields a remainder of 0 from the dividend, IS a factor of the dividend.

namely, if x⁴+3x³-6x+8 ÷ x+4, gives a remainder of 0, then x+4 IS a factor of x⁴+3x³-6x+8.
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Answer:

because x is the middle integer of three consecutive integers

=> The remaining 2 numbers respectively are x - 1 and x + 1

=> the sum of these three integers is

x - 1 + x + x + 1 = 3x

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The product of a number and 4 is three times the sum of that number and 6
maria [59]

Answer:

24?

Step-by-step explanation:

6 0
2 years ago
Kyra is using rectangular tiles of two types for a floor design. A tile of each type is shown below: Two rectangular tiles, rect
Rudiy27
Since there is no picture shown, I just plotted the given data points myself. That is shown in the attached picture. The blue rectangle is rectangle PQRS while the orange one is rectangle JKLM. I believe there are some choices for this question but you forgot to include. Nevertheless, I will give my observations from the given figure. 

The tile PQRS is bigger than tile JKLM. A rectangle is a two-dimensional shape that has two sets of equal parallel planes. Thus, its area is equal to the length multiplied by its width.

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8 0
3 years ago
Math again yay!...Ew math
Sliva [168]

Answer:

The graph of g(x) is wider.

Step-by-step explanation:

Parent function:

f(x)=x^2

New function:

g(x)=\left(\dfrac{1}{2}x\right)^2=\dfrac{1}{4}x^2

<u>Transformations</u>:

For a > 0

f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}

f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}

\begin{aligned} y =a\:f(x) \implies & f(x) \: \textsf{stretched/compressed vertically by a factor of}\:a\\ & \textsf{If }a > 1 \textsf{ it is stretched by a factor of}\: a\\  & \textsf{If }0 < a < 1 \textsf{ it is compressed by a factor of}\: a\\\end{aligned}

\begin{aligned} y=f(ax) \implies & f(x) \: \textsf{stretched/compressed horizontally by a factor of} \: a\\& \textsf{If }a > 1 \textsf{ it is compressed by a factor of}\: a\\  & \textsf{If }0 < a < 1 \textsf{ it is stretched by a factor of}\: a\\\end{aligned}

If the parent function is <u>shifted ¹/₄ unit up</u>:

\implies g(x)=x^2+\dfrac{1}{4}

If the parent function is <u>shifted ¹/₄ unit down</u>:

\implies g(x)=x^2-\dfrac{1}{4}

If the parent function is <u>compressed vertically</u> by a factor of ¹/₄:

\implies g(x)=\dfrac{1}{4}x^2

If the parent function is <u>stretched horizontally</u> by a factor of ¹/₂:

\implies g(x)=\left(\dfrac{1}{2}x\right)^2

Therefore, a vertical compression and a horizontal stretch mean that the graph of g(x) is wider.

4 0
1 year ago
The elimination method works when the equations are in what form
egoroff_w [7]
The answer is standard form. When the equation is set up as Ax + By = C where A and B are coefficients. 
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