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3 years ago

Right? Pls reply ASAP i need this before 11:45am in need pls hurry

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

6 0

Answer:

I think so.

Step-by-step explanation:

If you do the math 2(5-3) exponent 2 - 4 = 4

So if y is 4 you should be correct!

Thb I think you need to know the value of y to solve this question.

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I need help with this 2

Inessa [10]

Answer:

Ned lost 16 points total.

Step-by-step explanation:

If Ned got 6 questions wrong, and each one cost him four points, that would be 24 points total. 6x4=24. But since Ned got 8 additional points on a bonus question. It equals out to a total los of 18 points.

-24 + 8= -16

4 0

3 years ago

The accompanying diagram show a cross-section of a rectangular pyramid. The cross sectional area is 36in and is 3 inches from th

LiRa [457]

The area of the rectangular base is the amount of space on the rectangular base

The area of the rectangular base is 60 square inches

<h3>How to determine the area of the rectangular base?</h3>

The question is incomplete; as the diagram is not given.

So, I will apply the concept of similar shapes to determine the area of the rectangular base

To determine the area of the rectangular base, we make use of the following equivalent ratio:

Ratio = Height : Area

This gives

3 : 36 = 5 : Area

Express as fraction

36/3 = Area/5

Evaluate the quotient

12 = Area/5

Multiply both sides by 5

Area = 60

Hence, the area of the rectangular base is 60 square inches

Read more about areas at:

brainly.com/question/24487155

3 0

2 years ago

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

How many times does 12 go into 78

nirvana33 [79]

Answer: 6.5? I think this is the answer but not sure.

7 0

3 years ago

Read 2 more answers

I need too know what 2838x13 is right now or else ill flunk my test

alukav5142 [94]

Answer:

36894

Step-by-step explanation:

hope it helps

4 0

3 years ago

Read 2 more answers