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2 years ago

What is the step by step of 1/ cos x csc^2 x = sin x/ cot x

Mathematics

1 answer:

Effectus [21]2 years ago

7 0

Answer:

what is the step by step of 1/ cos x csc^2 x = sin x/ cot x

Step-by-step explanation:

csc

sin

cot

cos

sin

Given,

−

cos

sin

csc

−

cot

Simplify the right side.

csc

−

cot

sin

−

cos

sin

−

cos

sin

∴

, LS

RS.

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Whats 1 and 3 5ths times 2 and 3 4ths

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1 3/5 x 2 3/4
= 8/5 x 11/4
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= 4 2/5 (4 and 2/5)

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2 years ago

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Consider the equation below -2(bx -5 ) =16

gladu [14]

Answer:

bx = - 3

Step-by-step explanation:

- 2(bx - 5) = 16

- 2bx + 10 = 16

- 2bx + 10 - 10 = 16 - 10

- 2bx = 6

- 2bx ÷ - 2 = 6 ÷ - 2

bx = - 3

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2 years ago

Which allows you to temporarily use money that belongs to someone else?

MArishka [77]

Borrowing maybe is right

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2 years ago

There are 870 boys and 800 girls in a school

alina1380 [7]

Answer:

1060

Step-by-step explanation:

Given:

No. of boys = 870

No. of girls = 800

Probability that a boy chosen studies Spanish = $\dfrac{2}{3}$

Probability that a girl chosen studies Spanish = $\dfrac{3}{5}$

the number of boys in the school who study Spanish :

$\dfrac{2}{3}\times 870=290\times 2=580$

the number of girls in the school who study Spanish :

$\dfrac{3}{5}\times 800 \\=3\times 160\\\\=480$

Therefore, total number of students who study Spanish would be :

480+580=1060

8 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

2 years ago