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PIT_PIT [208]
3 years ago
13

may someone help me ASAP? I am solving polynomial equations and I don't know how to do it, here is the equation... x^4+11x^2=3x^

2+128
Mathematics
1 answer:
koban [17]3 years ago
8 0

<u>Answer:</u>

x=2\sqrt{2}, x=-2\sqrt{2}, x=4i, x=-4i

<u>Step-by-step explanation:</u>

x^4+11x^2=3x^2+128

First, we subtract 128 from both sides:

x^4+11x^2-128=3x^2

Then, we subtract 3x^2 from both sides:

x^4+8x^2-128=0

Rewrite the equation:

u^2+8u-128=0

Insert and solve:

u^2+8u-128=0\\u=8,u=-16\\x^2=8:x=2\sqrt{2},x= -2\sqrt{2} \\x^2=-16:x=4i,x=-4i

<em>Please give Brainliest</em>

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N is an integer.<br> Write the values of n such that -15 &lt; 3 ≤ 6<br><br> Help please ❤️❤️
Alex17521 [72]

Answer:

-4,-3,-2,-1,0,1

Step-by-step explanation:

First, doublepound and simplify it.

-15<3n

3n<6

Solve:

-5<n

n<2

Compound:

-5<n<2.

So the values are -4,-3,-2,-1,0,1

Hope this helps plz hit the crown :D

8 0
3 years ago
Read 2 more answers
QRINC offered new employees a starting salary of $34,862 in 2013. What would a comparable starting salary have been in 2003?
Vlad1618 [11]

Complete Question

Table of Annual CPI values

2003-184.00            

2004-188.90

2005-195.3

2006-201.6

2007-207.342

2008-215.303

2009-214.537

2010-218.056

2011-224.939

2012-229.594

2013-232.957

2014-236.736

QRINC offered new employees a starting salary of $34,862 in 2013. What would a comparable starting salary have been in 2003?

Answer:

C=27535.5881128

Step-by-step explanation:

From the question we are told that

CPI for 2003(index)=2003-184.00  

CPI for 2013(index)=2013-232.957  

Starting salary in 2013 at $34,862

Generally  comparable starting salary C is given as

   C=\frac{starting salary}{CPI for 2013(index} *CPI for 2003(index)

   C=\frac{34,862}{232.957} *184

Therefore C the comparable starting salary is givrn to be

   C=27535.5881128

   C=27536 appro

8 0
3 years ago
Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7
rosijanka [135]
\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx

Notice that x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1). Decompose the integrand into partial fractions:

\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}
=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}

Integrating term-by-term, you get

\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx
=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C
5 0
3 years ago
The following table gives results from two groups of students who took a nonproctored test. Use a 0.01 significance level to tes
Nana76 [90]

Answer:

Original claim is \mu_{1} =\mu_{2}

Opposite claim is \mu_{1} \neq \mu_{2}

Null and alternative hypotheses:

H_{0}:\mu_{1} = \mu_{2}

H_{1} : \mu_{1} \neq \mu_{2}

Significance level: 0.01

Test statistic:

We can use TI-84 calculator to find the test statistic and P-value. The steps are as follows:

Press STAT and the scroll right to TESTS

Scroll down to 2-SampTTest... and scroll to stats.

Enter below information.

\bar{x_{1}}=70.29

Sx1=22.09

n_{1} = 30

\bar{x_{2}}=74.26

Sx2=18.15

n_{2} = 32

\mu_{1} \neq \mu_{2}

Pooled: Yes

Calculate.

The output is in the attachment.

Therefore, the test statistic is:

t=-0.78

P-value: 0.4412

Reject or fail to reject: Fail to reject

Final Conclusion: Since the p-value is greater than the significance level, we, therefore, fail to reject the null hypothesis and conclude that the there is sufficient evidence to support the claim that the samples are from populations with the same mean.

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3 years ago
2(3x – 1) = –6x – 2 can someone help
Gnom [1K]

Let's solve your equation step-by-step.

2(3x−1)=−6x−2

Step 1: Simplify both sides of the equation.

2(3x−1)=−6x−2

(2)(3x)+(2)(−1)=−6x+−2(Distribute)

6x+−2=−6x+−2

6x−2=−6x−2

Step 2: Add 6x to both sides.

6x−2+6x=−6x−2+6x

12x−2=−2

Step 3: Add 2 to both sides.

12x−2+2=−2+2

12x=0

Step 4: Divide both sides by 12.

12x/12= 0/12

x=0

Answer:

x=0

HOPE THIS HELPS!!! :)

5 0
3 years ago
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