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2 years ago

PLEASE HELP IS THIS CORRECT GUYS YES OR NO!

Mathematics

1 answer:

mote1985 [20]2 years ago

4 0

Answer:

yes

Step-by-step explanation:

i think if it's wrong pls don't hate me

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Can you pls help me the quiz is timed noo links and can you give me an explanation

Shtirlitz [24]

Answer:

2 and 96 (lowest and highest #s far from any other #s listed

7 0

3 years ago

Solve for y 1/3x+1/4y=2

Zigmanuir [339]

Answer:

$y=\frac{-4}{3} x+8$

Step-by-step explanation:

Add -1/3x to both sides.

$\frac{1}{4}y=\frac{-1}{3}+2$

Divide both sides by 1/4.

$y=\frac{-4}{3} x+8$

hope this helps

6 0

3 years ago

Read 2 more answers

Which expression means “ multiply 3 by the sum of m and p “

JulijaS [17]

The answer would be A. hope this helps

6 0

2 years ago

81 POINTS

Jobisdone [24]

Base Case: plug in n = 1 (the smallest positive integer)

If n = 1, then 3n-2 = 3*1-2 = 1. Square this and we see that (3n-2)^2 = 1^2 = 1

On the right hand side, plugging in n = 1 leads to...

n*(6n^2-3n-1)/2 = 1*(6*1^2-3*1-1)/2 = 1

Both sides are 1. So that confirms the base case.

-------------------------------

Inductive Step: Assume that

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

is a true statement for some positive integer k. If we can show the statement leads to the (k+1)th case being true as well, then we will have sufficiently proven the overall statement to be true by induction.

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 + (3(k+1)-2)^2 = (k+1)*(6(k+1)^2-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3(k+1)-2)^2 = (k+1)*(6(k^2+2k+1)-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3k+3-2)^2 = (k+1)*(6k^2+12k+6-3k-3-1)/2

k*(6k^2-3k-1)/2 + (3k+1)^2 = (k+1)*(6k^2+9k+2)/2

k*(6k^2-3k-1)/2 + 9k^2+6k+1 = (k+1)*(6k^2+9k+2)/2

(6k^3-3k^2-k)/2 + 2(9k^2+6k+1)/2 = (k*(6k^2+9k+2)+1(6k^2+9k+2))/2

(6k^3-3k^2-k + 2(9k^2+6k+1))/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3-3k^2-k + 18k^2+12k+2)/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3+15k^2+11k+2)/2 = (6k^3+15k^2+11k+2)/2

Both sides simplify to the same expression, so that proves the (k+1)th case immediately follows from the kth case

That wraps up the inductive step. The full induction proof is done at this point.

7 0

3 years ago

Please help me with this. simplify and show all your work.

iren2701 [21]

Let's take this variable by variable. z to the power of 4 over z to the power of one would equal z to the power of 3. At this point, there is no longer a z variable in the denominator.
X to the power of 2 over x to the power of 1 would equal x. At this point there is also no x variable in the denominator.
Y to the power of 1 over y to the power of 2 would equal (1/y). At this point, there is no y in the NUMERATOR.

Final Answer: $\frac{ z^{3}x }{y}$

5 0

3 years ago