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3 years ago

W cubed x W (-9th Power)

Mathematics

1 answer:

aalyn [17]3 years ago

3 0

Answer:

$\frac{1}{w {}^{6} }$

Step-by-step explanation:

${w}^{3} \times w {}^{ - 9}$

equal

$w {}^{ - 6}$

Using the product of power rule,

${x}^{a } \times x {}^{e} = x {}^{a + e}$

Then apply negative power rule

$a {}^{ - x} = \frac{1}{a {}^{x} }$

So it now becomes

$\frac{1}{w {}^{6} }$

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So if Caelyn collected 21 shells in total and her brother collected 4 times as much all you need to do is multiply 21 by 4 and you get your answer which is 84

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4 years ago

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Mars2501 [29]

Answer:

a/(bxy) if b≠0, x≠0, y≠0, x≠y

Step-by-step explanation:

You have correctly simplified the expression. The conditions placed on both the original and simplified forms are that none of the denominator factors may be zero. The denominator factors are b, x, y, and (x-y).

$\dfrac{a}{bxy}\text{ if $b\ne0;$ $x\ne0;$ $y\ne0;$ $x\ne y$}$

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2 years ago

Find two numbers whose sum is 55 and whose product is 646

Svet_ta [14]

Answer:

17 and 38

Step-by-step explanation:

x + y = 55......x = 55 - y

x * y = 646

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3 years ago

Will crown the correct answers brainliest 3x

Solnce55 [7]

Answer :

(1) $\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) $\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{15xy}$

(3) $\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{14xy}$

(4) $\frac{2}{5x}+\frac{3}{7y}=\frac{6y+15x}{21xy}$

(5) $\frac{7}{11x}-\frac{1}{33y}=\frac{231y-11x}{363xy}$

(6) $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{7y+42y-2x}{14xy}$

Step-by-step explanation :

(1) The given expression is: $\frac{1}{x}+\frac{1}{y}$

$\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) The given expression is: $\frac{1}{5x}+\frac{1}{3y}$

$\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}$

(3) The given expression is: $\frac{1}{7x}-\frac{1}{2y}$

$\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}$

(4) The given expression is: $\frac{2}{5x}+\frac{3}{7y}$

$\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}$

(5) The given expression is: $\frac{7}{11x}-\frac{1}{33y}$

$\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}$

(6) The given expression is: $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}$

$\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}$

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3 years ago

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Answer:

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