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2 years ago

Can someone please help me ?

Mathematics

1 answer:

12345 [234]2 years ago

8 0

I won’t be able to answer unless i have points to go off of

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Help meeeeeeee please :(.

vodomira [7]

Answer:

$x^{3} -5x^{2} -6x +8 \frac{-2}{x + 3}$

Step-by-step explanation:

$x^{3} -5x^{2} -6x +8 \frac{-2}{x + 3}$

x + 3 / $x^{4} - 2x^{3}-21x^{2} -10x +22$

-( $x^{4} + 3x^{3}$ )

- $5x^{3} -21x^{2} -10x +22$

-( $-5x^{3} -15x^{2}$ )

$-6x^{2} -10x +22$

-( $-6x^{2} -18x$ )

8x +22

-(8x + 24)

$\frac{-2}{x + 3}$

4 0

3 years ago

PLEASE help me I beg you

Likurg_2 [28]

Answer:

pls, don't do that

Step-by-step explanation:

3 0

2 years ago

pls help will give brainliest Rena used the steps below to evaluate the expression (StartFraction (x Superscript negative 3 Base

kaheart [24]

Answer:

First "order of operations" mistake: step 2

First arithmetic mistake: step 4

Step-by-step explanation:

As we understand Rena's work, she wants to simplify ...

$\left(\dfrac{x^{-3}y^{-2}}{2x^4y^{-4}}\right)^{-3}$

for x = -1 and y = 2.

Her work seems to be ...

Step 1

$\text{Substitute $x=-1$ and $y=2$ into the expression}\\\\\left(\dfrac{(-1)^{-3}2^{-2}}{2(-1)^42^{-4}}\right)^{-3}\qquad\text{no error}$

Step 2

$\text{Simplify the parentheses}\\\\\left(\dfrac{2^4}{2(-1)^4(-1)^32^2}\right)^{-3}=\left(\dfrac{2^2}{2(-1)^7}\right)^{-3}\qquad\text{order of operations error}$

Step 3

$\text{Evaluate the power to a power}\\\\\dfrac{2^{-6}}{2^{-3}(-1)^{21}}\qquad\text{no error}$

Step 4

$\text{Use reciprocals and find the value}\\\\\dfrac{1}{2^32^6(-1)^{21}}=\dfrac{1}{8\cdot 64\cdot (-1)}=\dfrac{-1}{512}\qquad\text{error: $2^3$ is used instead of $2^{-3}$}$

_____

So, the first arithmetic error is in Step 4. However, the order of operations requires exponents be evaluated first. Doing that makes step 2 look like ...

$\left(\dfrac{-\dfrac{1}{4}}{2(1)\dfrac{1}{16}}\right)^{-3}=(-2)^{-3}\qquad\text{proper Step 2}$