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lbvjy [14]
3 years ago
15

What is the value of x in the equation 7x+2v =48 when y =3​

Mathematics
2 answers:
Phantasy [73]3 years ago
4 0

Answer:

3 3/7

Step-by-step explanation:

7x+2v=48

2v. -2

-------

46

7x+46=48

-46. -46

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

7x=2/7. x=2/7=1/7

x=1/7

bonufazy [111]3 years ago
4 0

Answer:

x = 6

Step-by-step explanation:

7x+2y =48 when y =3

Substitute for y in the equation.

7x + 2(3) = 48

7x + 6 = 48

7x = 48 - 6

7x = 42

x = 42/7

x = 6

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What is 15/6 in mixed number
jenyasd209 [6]
15/6 is an improper fraction.

Lets convert this into a proper fraction.

How many times does 6 go into 15?

6*2 = 12

Only twice, so our whole number is 2.

15-12 = 3

3 is our numerator.

2   3/6  =   2  1/2

Mixed number form:   2   1/2

Final answer: 2  1/2
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What is the word form for the number 37,917,545??? <br> i am in elementry
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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
In triangle ABC, angles A and angle B are complementary, where cos A = 0.5.
Valentin [98]

Answer:

<h3>∴∠B=30°</h3>

Step-by-step explanation:

Complementary:

Two angles is said to complementary when the sum of two angles is 90 degree.

Given that,

∠A and ∠B are complementary.

Then,  ∠A + ∠B = 90°

Cos A = 0.5

\Rightarrow A=cos^{-1}(0.5)

\Rightarrow A=cos^{-1}(cos \ 60^\circ)

\Rightarrow A=60^\circ

Since

 ∠A + ∠B = 90°

⇒ 60°+ ∠B = 90°      [ plug ∠A= 60°]

⇒∠B=90°-60°

⇒∠B=30°

<h3>∴∠B=30°</h3>
3 0
3 years ago
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