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2 years ago

1. Suppose that cos B = 0.68. What is the value of sin(a)? Explain.

Mathematics

1 answer:

Papessa [141]2 years ago

5 0

The value of Sin(a) suppose that Cos B = 0.68 in which case, a + B = 90° is; 0.68.

<h3>Trigonometry</h3>

First, when we have two angles, whose sum equals, 90°.

In essence, Since the algebraic sum of angles A and B is 90°;

A + B = 90.

B = 90 -A.

By trigonometric identity;

Sin A = Cos (90 - A)

Where; (90 - A) = B.

Therefore; Sin A = Cos B = 0.68.

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Step-by-step explanation:

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2 years ago

Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

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$\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

ie, $5^\frac{\bf -5}{\bf 6}$

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

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ch4aika [34]

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