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GenaCL600 [577]
3 years ago
8

Last year, Mr. Jones made $30,000. His boss just informed him that he will be receiving at least an 11.2% raise for this year. H

ow much will he make this year?
Mathematics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

$33 336

Step-by-step explanation:

Increase = 30 000 × 0.112

Increase = $3336

New salary = 30 000 + 3336

New salary = $33 336

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A triangle has side lengths 3.4 cm, 4 cm, and 5.2 cm. The angle measures are 40°, 50°, and 90°.
igomit [66]

Answer:

Right scalene

7 0
3 years ago
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5x + 4x +1 need answer for this can anyone help?
nalin [4]
9x+1 is the answer.
I hope this helped
8 0
3 years ago
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Please help. Im dying for the explanation. I will mark you brainalist.
AlekseyPX

Step-by-step explanation:

Given that the mean of the numbers is 12

mean =  \frac{sum \:  \: of \: observations}{total \: number \: of \: observations}

so here

let's that one as x and another one as 2x since it is given that one is twice the other

Therefore we get

mean =  \frac{13 + x + 4 + 16 + 6 + 2x}{6}

Given mean = 12

12 =  \frac{39 + 3x}{6}

72 = 39 + 3x

3x = 33

x = 11

Hence it's 11 and 22

5 0
2 years ago
An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 

substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
so the maximum value =\frac{1}{3}
minimum value =- \frac{1}{3}


4 0
3 years ago
What are potential roots of p(x) = x4 + 22x2 – 16x – 12
kow [346]
Simplify <span>22\times 2<span>22×2</span></span> to <span>44<span>44</span></span>
<span><span>{x}^{4}+44-16x-12<span><span>x<span><span>​4</span><span>​​</span></span></span>+44−16x−12</span></span>Collect like terms
<span><span>{x}^{4}+(44-12)-16x<span><span>x<span><span>​4</span><span>​​</span></span></span>+(44−12)−16x</span></span> Simplify</span><span><span>{x}^{4}+32-16x<span><span>x<span><span>​4</span><span>​​</span></span></span>+32−16x</span></span><span>
</span></span></span>
8 0
3 years ago
Read 2 more answers
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