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3 years ago

Write this number in expanded form.4,508,227

Mathematics

2 answers:

Step2247 [10]3 years ago

8 0

4,000,000+500,000+8,000+200+20+7

Semmy [17]3 years ago

7 0

4000000+500000+8000+200+20+7

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Round 2.496 to the nearest hundredth

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Answer:2.50

Step-by-step explanation:

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3 years ago

Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Drupady [299]

Answer:

The equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

Let us solve the expression step by step to get the equivalent

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

$\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=x^{2\cdot \frac{1}{7}}$

$=x^{\frac{2}{7}}$

also

$\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=y^{3\cdot \frac{1}{5}}$

$=y^{\frac{3}{5}}$

so the expression becomes

$\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$ ∵ $\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}$

Thus, the equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

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3 years ago

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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4 years ago

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