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4vir4ik [10]
3 years ago
5

Algebra 2 help please.

Mathematics
1 answer:
Oksanka [162]3 years ago
4 0
\frac{\sqrt{4^{3}}\sqrt[3]{x}\sqrt{y}}{\sqrt{z}} * \frac{\sqrt[5]{y^{2}}\sqrt[3]{z}}{\sqrt{x^{4}}} * \frac{\sqrt[5]{x^{3}}\sqrt[4]{x^{6}}}{\sqrt[3]{y^{2}}}
\frac{\sqrt{64}\sqrt[3]{x}\sqrt{y}}{\sqrt{z}} * \frac{\sqrt[5]{y^{2}}\sqrt[3]{z}}{x^{2}} * \frac{x^{2}\sqrt[5]{x^{3}}\sqrt[4]{x^{2}}}{\sqrt[3]{y^{2}}}
\frac{8\sqrt[3]{x}\sqrt{y}}{\sqrt{z}} * \frac{\sqrt[5]{y^{2}}\sqrt[3]{z}}{x^{2}} * \frac{x^{2}\sqrt[5]{x^{3}}\sqrt[4]{x^{2}}}{\sqrt[3]{y^{2}}}
\frac{8\sqrt{y}\sqrt[3]{xz}\sqrt[4]{x^{2}}\sqrt[5]{x^{3}y^{2}}}{\sqrt{z}\sqrt[3]{y^{2}}}
\frac{8\sqrt{yz}\sqrt[3]{xy^{2}z}\sqrt[4]{x^{2}}\sqrt[5]{x^{3}y^{2}}}{y^{2}z}
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<u>Janet ⇒ 3 postcards ⇒ $1.05 </u>

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<u>We see in all 3 cases cost of postcard is same, the cost and numbers are proportional and the relationship is:</u>

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