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Fantom [35]
3 years ago
5

Solve by adding and subtracting rational expressions. (3x)/(5xy^(4))+(6x^(2))/(15x^(2)y) Remember to show your work!

Mathematics
1 answer:
slamgirl [31]3 years ago
4 0

Answer:

Step-by-step explanation:

\frac{3x}{5xy^{4}}+\frac{6x^{2}}{15x^{2}y}=\frac{3}{5y^{4}}+\frac{6}{15y}\\\\=\frac{3}{5y^{4}}+\frac{2}{5y}\\\\=\frac{3}{5y^{4}}+\frac{2*y^{3}}{5y*y^{3}}\\\\=\frac{3}{5y^{4}}+\frac{2y^{3}}{5y^{4}}\\\\=\frac{3+2y^{3}}{5y^{4}}

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Answer:

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Step-by-step explanation:

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3 years ago
2. A square-based tent has the cross-sectional
ollegr [7]

(a) Length of the height is 2.732 m

(b) Length of the base is 5.466 m

<u>Explanation:</u>

An image is attached for reference.

(a)

In ΔAOB,

sin 30^o = \frac{AO}{AB} \\\\0.5 = \frac{AO}{2} \\\\AO = 1 m

In ΔBGD,

sin 60^o = \frac{BG}{BD} \\\\0.866 = \frac{BG}{2} \\\\BG = 1.732 m

According to the figure, BG = OE = 1.732 m

Height of the tent, AE = AO + OE

                                  = 1 m + 1.732 m

                                  = 2.732 m

(b)

DF = ?

In ΔAOB,

tan 30^o = \frac{AO}{OB} \\\\0.577 = \frac{1}{OB} \\\\OB = 1.733 m\\\\\\

According to the figure, OB = GE = 1.733 m

In ΔBGD,

tan 60^o = \frac{BG}{DG} \\\\1.732 = \frac{1.732}{DG}\\ \\DG = 1m

According to the figure, DE = DG + GE

                                      DE = 1 m + 1.733 m

                                     DE = 2.733 m

Length of the base, DF = 2 X DE

                              DF = 2 X 2.733 m

                               DF = 5.466 m

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Answer:

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Step-by-step explanation:

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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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