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scoray [572]
2 years ago
7

(GIVING BRAINLIEST!!)

Mathematics
1 answer:
BartSMP [9]2 years ago
8 0

Answer:

1/9

Step-by-step explanation:

Divide 1/3 divided by 3 (1/9)

Keep, Change, Flip

(Keep the first fraction, change the division sign to multiplication sign, flip the last fraction)

The equation would be

1/3*1/3=1/9

Hope I helped!! :)

Pls mark brainliest if correct!!!

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kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
Each unit on a coordinate plane represents 1 mile. One end of the road starts at (212,23). The road ends at (6,23). How long is
olya-2409 [2.1K]
Distance formula : sqrt (x2 - x1)^2 + (y2 - y1)^2
(212,23)...x1 = 212 and y1 = 23
(6,23)......x2 = 6 and y2 = 23
now we sub
d = sqrt ((6 - 212)^2 + (23 - 23)^2)
d = sqrt ((- 206)^2 + (0^2))
d = sqrt (42436 + 0)
d = sqrt 42436
d = 206 miles <==
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Question 5 is attached :)
OverLord2011 [107]
I hope this helps you

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LekaFEV [45]

Answer:

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Step-by-step explanation:

Please

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