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3 years ago

The measures of two complementary angles are m angle 1 = (10x+7) and m angle 2=(9x-12) find the measures of both angles

Mathematics

1 answer:

USPshnik [31]3 years ago

5 0

Answer:

m<1 = 57°

m<2 = 33°

Step-by-step explanation:

To find the numerical measure of both angles, let's come up with an equation to determine the value of x.

Given that m<1 = (10x +7)°, and m<2 = (9x - 12)°, where both are complementary angles, therefore, it means, both angles will add up to give us 90°.

Equation we can generate from this, is as follows:

(10x + 7)° + (9x - 12)° = 90°

Solve for x

10x + 7 + 9x - 12 = 90

Combine like terms

19x - 5 = 90

Add 5 to both sides

19x = 90 + 5 (addition property not equality)

19x = 95

Divide both sides by 19

x = 5

m<1 = (10x +7)°

Replace x with 5

m<1 = 10(5) + 7 = 50 + 7 = 57°

m<2 = (9x - 12)

Replace x with 5

m<2 = 9(5) - 12 = 45 - 12 = 33°

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dalvyx [7]

Answer:

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Step-by-step explanation:

We have:

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So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

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$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

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We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

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Step-by-step explanation:

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You know that:

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