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bonufazy [111]
3 years ago
15

Identifying Additive Inverses

Mathematics
1 answer:
mestny [16]3 years ago
4 0

Answer:

he additive inverse of:

a)  -6x^2-x-2 is : 6x^2+x+2

b)  6x^2-x+2  is :  -6x^2+x-2

c)  6x^2+x-2  is :  -6x^2-x+2

d)  6x^2+x+2  is :  -6x^2-x-2

Step-by-step explanation:

You need to consider that the additive inverse of a polynomial is that polynomial that consists of the opposite of each term of the polynomial given.

Then, the additive inverse of:

a)  -6x^2-x-2 is :  6x^2+x+2

b)  6x^2-x+2  is :  -6x^2+x-2

c)  6x^2+x-2  is :  -6x^2-x+2

d)  6x^2+x+2  is :  -6x^2-x-2

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Plsss help I don’t have time!!!!!
Elis [28]

Answer:

D) y=5x+20

Step-by-step explanation:

8 0
2 years ago
Using the curve of best fit, what would be the output (y) when the input (x) is 7?
IrinaVladis [17]

Answer:

y = 260

Step-by-step explanation:

locate x = 7 on the x- axis, go vertically up to meet the graph at (7, 260 )

when input is 7 then output y = 260

4 0
2 years ago
2<br> .<br> 2<br> 12n = 42<br> NEED HELP ASAP⬆️
Anni [7]

Answer:

  6  

 —————

 n + 8

Step-by-step explanation:

Step by Step Solution:

More Icon

STEP

1

:

Equation at the end of step 1

 ((12•(n3))-(24•(n2)))       (12n-42)      

 —————————————————————•———————————————————

  (((4•(n2))-22n)+28)  ((6•(n3))+(24•3n2))  

STEP  

2

:

Equation at the end of step

2

:

 ((12•(n3))-(24•(n2)))      (12n-42)      

 —————————————————————•——————————————————

  (((4•(n2))-22n)+28)  ((2•3n3)+(24•3n2))  

STEP

3

:

            12n - 42  

Simplify   ——————————

           6n3 + 48n2

STEP

4

:

Pulling out like terms

4.1     Pull out like factors :

  12n - 42  =   6 • (2n - 7)  

STEP

5

:

Pulling out like terms

5.1     Pull out like factors :

  6n3 + 48n2  =   6n2 • (n + 8)  

Equation at the end of step

5

:

 ((12•(n3))-(24•(n2)))  (2n-7)  

 —————————————————————•————————

  (((4•(n2))-22n)+28)  n2•(n+8)

STEP  

6

:

Equation at the end of step

6

:

 ((12•(n3))-(24•(n2)))  (2n-7)  

 —————————————————————•————————

    ((22n2-22n)+28)    n2•(n+8)

STEP  

7

:

Equation at the end of step

7

:

 ((12•(n3))-(23•3n2))  (2n-7)  

 ————————————————————•————————

     (4n2-22n+28)     n2•(n+8)

STEP  

8

:

Equation at the end of step

8

:

 ((22•3n3) - (23•3n2))      (2n - 7)  

 ————————————————————— • ————————————

   (4n2 - 22n + 28)      n2 • (n + 8)

STEP

9

:

             12n3 - 24n2  

Simplify   ——————————————

           4n2 - 22n + 28

STEP

10

:

Pulling out like terms

10.1     Pull out like factors :

  12n3 - 24n2  =   12n2 • (n - 2)  

STEP

11

:

Pulling out like terms

11.1     Pull out like factors :

  4n2 - 22n + 28  =   2 • (2n2 - 11n + 14)  

Trying to factor by splitting the middle term

11.2     Factoring  2n2 - 11n + 14  

The first term is,  2n2  its coefficient is  2 .

The middle term is,  -11n  its coefficient is  -11 .

The last term, "the constant", is  +14  

Step-1 : Multiply the coefficient of the first term by the constant   2 • 14 = 28  

Step-2 : Find two factors of  28  whose sum equals the coefficient of the middle term, which is   -11 .

     -28    +    -1    =    -29  

     -14    +    -2    =    -16  

     -7    +    -4    =    -11    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -7  and  -4  

                    2n2 - 7n - 4n - 14

Step-4 : Add up the first 2 terms, pulling out like factors :

                   n • (2n-7)

             Add up the last 2 terms, pulling out common factors :

                   2 • (2n-7)

Step-5 : Add up the four terms of step 4 :

                   (n-2)  •  (2n-7)

            Which is the desired factorization

Canceling Out :

11.3    Cancel out  (n-2)  which appears on both sides of the fraction line.

Equation at the end of step

11

:

   6n2      (2n - 7)  

 —————— • ————————————

 2n - 7   n2 • (n + 8)

STEP

12

:

Canceling Out

12.1    Cancel out  (2n-7)  which appears on both sides of the fraction line.

Canceling Out :

12.2    Canceling out n2 as it appears on both sides of the fraction line

Final result :

   6  

 —————

 n + 8

8 0
3 years ago
The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi
eimsori [14]

Answer:

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=4.2 represent the sample mean  

s=1.4 represent the sample standard deviation  

n=16 represent the sample selected  

\alpha=0.05 significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:\mu = 3.6    

Alternative hypothesis:\mu \neq 3.6    

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with df=n-1=16-1=15 degrees of freedom. The value for \alpha=0.05 and \alpha/2=0.025 so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got t_{crit}=\pm 2.131    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

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