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kherson [118]
3 years ago
13

10m3 of air weigh 13 kg. The weight of air is directly proportional to its volume. Find the constant of variation; it is called

density
Mathematics
1 answer:
love history [14]3 years ago
8 0
\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\
\textit{weight of air is directly proportional to its volume}\qquad W=kV
\\\\\\
\textit{we also know that }
\begin{cases}
V=10\\
W=13
\end{cases}\implies 13=k10\implies \cfrac{13}{10}=k
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4 consecutive odd numbers....x, x + 2, x + 4, x + 6

2(x + x + 2) = 3(x + 4 + x + 6) - 40
2(2x + 2) = 3(2x + 10) - 40
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x = -14/-2
x = 7

x + 2 = 7 + 2 = 9
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Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-
Charra [1.4K]

Answer:

The statement is true is for any n\in \mathbb{N}.

Step-by-step explanation:

First, we check the identity for n = 1:

(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}

1 = \frac{1\cdot 1\cdot 3}{3}

1 = 1

The statement is true for n = 1.

Then, we have to check that identity is true for n = k+1, under the assumption that n = k is true:

(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)

(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)

k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)

2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)

(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)

Therefore, the statement is true for any n\in \mathbb{N}.

4 0
3 years ago
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