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vovangra [49]
4 years ago
13

In the triangle below, Four-fifths represents which ratio?

Mathematics
1 answer:
vovangra [49]4 years ago
8 0

Answer:

(A) sin C =\frac{4}{5}

Step-by-step explanation:

Opposite,AB = 4,

Hypotenuse,BC = 5

Adjacent, AC = 3.

sin \theta =\frac{opposite}{Hypotenuse}\\ sin C =\frac{4}{5}

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find the equation of the straight line which has a gradient of 3 and passes through the point (-1, 3)
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Read 2 more answers
The answer is red show step by step of the pro tem to get the answer
miskamm [114]

In order to rationalize the denominator of each expression, we need to multiply the expression by the same radical in the denominator, this way we can remove the radical from the denominator.

9)

\frac{5\sqrt{4}}{\sqrt{3}}(\cdot\sqrt{3})=\frac{5\sqrt{4}\sqrt{3}}{(\sqrt{3})^2}=\frac{5\cdot2\cdot\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}

10)

-\frac{5}{3\sqrt{2}}(\cdot\sqrt{2})=-\frac{5\sqrt{2}}{3(\sqrt{2})^2}=-\frac{5\sqrt{2}}{3\cdot2}=-\frac{5\sqrt{2}}{6}

11)

\frac{2\sqrt{3}}{4\sqrt{5}}(\cdot\sqrt{5})=\frac{2\sqrt{3}\sqrt{5}}{4(\sqrt{5})^2}=\frac{2\sqrt{15}}{4\cdot5}=\frac{\sqrt{15}}{2\cdot5}=\frac{\sqrt{15}}{10}

12)

\frac{\sqrt{5}}{\sqrt{2}}(\cdot\sqrt{2})=\frac{\sqrt{5}\sqrt{2}}{(\sqrt{2})^2}=\frac{\sqrt{10}}{2}

3 0
1 year ago
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