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3 years ago

Pls help! 16 points:)

Mathematics

2 answers:

kenny6666 [7]3 years ago

7 0

Answer: 52 degrees

Step-by-step explanation:

most likley like if correct

siniylev [52]3 years ago

4 0

The correct answer is 52 degrees

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Ike Phillips worked 6.7 hours at time-and-a-half pay and 3.4

ikadub [295]

Working out the different pay rates earned by Ike Phillips :

Time and half pay rate = $17.055
Double pay rate = $22.74
Time and half earning = $114.2685
Double rate earning = $77.316
Gross earning = $646.3845

Let :

Regular rate = $11.37

Total earning for the week = $454.80

Double pay hours = 3.4

Time and half pay hours = 6.7

Time and half pay rate = 1.5 × regular pay rate = 1.5 × 11.37 = $17.055

Double pay rate = 2 × regular pay rate = 2 × 11.37 = $22.74

Time and half earning = rate × time = $17.055 × 6.7 = $114.2685

Double rate earning = rate × time = $22.74 × 3.4 = $77.316

Gross earning = (regular + time and half + double earning) = $(454.80+114.2685+77.316) = $646.3845

Therefore, Ike's gross earning for the week is $646.3845

Learn more : brainly.com/question/24635246

4 0

3 years ago

Plz factor this completely

MrMuchimi

What am I supposed to be factoring?

3 0

3 years ago

Read 2 more answers

Pls help I will mark brainlist if correct !! :)

Zielflug [23.3K]

Answer:

the first one is correct

Step-by-step explanation:

mark brainliest please

4 0

3 years ago

Read 2 more answers

Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.

nignag [31]

Answer:

we need to prove : for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

1) check divisibility for n=1, $f(1)=(1)^{5}-1=0$ (divisible)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

3) Induction,

$f(k+1)=(k+1)^{5}-(k+1)$

$=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1$

$=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k$

Now, $f(k+1)-f(k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k$

$f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k$

Take out the common factor,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

add both the sides by f(k)

$f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)$

We have proved that difference between $f(k+1)$ and $f(k)$ is divisible by 5.

so, our assumption in step 2 is correct.

Since $f(k)$ is divisible by 5, then $f(k+1)$ must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

3 0

3 years ago

Be sure to use a comma after three please value of one need it 945×1000

JulijaS [17]

Answer:

94,5000

Step-by-step explanation:

7 0

3 years ago