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3 years ago

Someone pls also include any work if they can thank you! I’ll mark brainliest!

Mathematics

1 answer:

zysi [14]3 years ago

5 0

Answer/Step-by-step explanation:

A1. Reference angle = 41°

Hypotenuse = x

Adjacent = 7

Thus, applying trigonometric ratio, we would have:

Cos 41 = 7/x

Multiply both sides by x

x × cos 41 = 7

Divide both sides by cos 41

x = 7/cos 41

x = 9.3 (nearest tenth)

A2. Reference angle = 65°

Hypotenuse = 6

Opposite = y

Thus, applying trigonometric ratio, we would have:

Sin 65 = y/6

Multiply both sides by 6

6 × sin 65 = y

5.4 = y (nearest tenth)

y = 5.4

A3. Reference angle = 50°

Hypotenuse = z

Opposite = 8

Thus, applying trigonometric ratio, we would have:

Sin 50 = 8/z

Multiply both sides by z

z × sin 50 = 8

Divide both sides by sin 50

z = 8/sin 50

z = 10.4 (nearest tenth)

B1. Reference angle = 49°

Hypotenuse = x

Opposite = 7

Thus, applying trigonometric ratio, we would have:

Sin 49 = 7/x

Multiply both sides by x

x × sin 49 = 7

Divide both sides by sin 49

x = 7/sin 49

x = 9.3 (nearest tenth)

B2. Reference angle = 25°

Hypotenuse = 6

Adjacent = y

Thus, applying trigonometric ratio, we would have:

Cos 25 = y/6

Multiply both sides by 6

6 × cos 25 = y

y = 5.4 (nearest tenth)

B3. Reference angle = 40°

Hypotenuse = z

Adjacent = 8

Thus, applying trigonometric ratio, we would have:

Cos 40 = 8/z

Multiply both sides by z

z × cos 40 = 8

Divide both sides by cos 40

z = 8/cos 40

z = 10.4 (nearest tenth)

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3 years ago

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$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
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