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3 years ago

-2/7х=-1/3, срочно(если что это дроби)

Mathematics

1 answer:

maria [59]3 years ago

7 0

Exact form: x=7/6

Decimal form: x=1.16

Mixed number form: x=1 1/6

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6 hours

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3 years ago

There are 150 children at an art camp. 71 signed up for painting. 62 children signed up for sculpting and 28 of them also signed

ki77a [65]

Answer:

99/150= 62/150

Step-by-step explanation:

6 0

3 years ago

Can you help me with number 10

rewona [7]

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{(3^{-2})^2}{3^{-3}}\implies \cfrac{3^{-2\cdot 2}}{3^{-3}}\implies \cfrac{3^{-4}}{3^{-3}}\implies \cfrac{1}{3^{-3}\cdot 3^4}\implies \cfrac{1}{3^{-3+4}}\implies \cfrac{1}{3^1}\implies \cfrac{1}{3}$

4 0

3 years ago

How can i prove this property to be true for all values of n, using mathematical induction.

chubhunter [2.5K]

Proof -

So, in the first part we'll verify by taking n = 1.

$\implies \: 1 = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6}$

$\implies{ \frac{1(2)(3)}{6} }$

$\implies{ 1}$

Therefore, it is true for the first part.

In the second part we will assume that,

$\: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }$

and we will prove that,

$\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}$

$\: {{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 2) (2k + 3)}{6}}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}$

Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.

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8 0

2 years ago

What’s the square root of 108 a to the 7th power b to the 5th power

MissTica

Answer:

Step-by-step explanation:

Factor all the factors making up the square root.

108: 2*2*3*3*3

a^7: a*a*a*a*a*a*a

b^5:b * b * b * b * b

The rule for taking a square root of this is

For every pair, you get to take one number outside the root sign and throw the other one away.

2*3*a * a * a* b * b√(3*a * b)

6a^3b^2 * √(3*a*b)

4 0

3 years ago