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2 years ago

What is cot x(sin^2x+cos^2x)

Mathematics

1 answer:

shusha [124]2 years ago

6 0

Answer:

cot(x)

Step-by-step explanation:

$cot\theta(sin^2\theta+cos^2\theta)$

$cot\theta(1)$

$cot\theta$

Recall the Pythagorean Identity $sin^2x+cos^2x=1$

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You buy a new laptop for $300$300. The sales tax is 6%6%. What is the total cost for the laptop including the sales tax?

cricket20 [7]

Answer:

Cost Including tax would be $318

Step-by-step explanation:

Given:

Price of laptop = $300

Sales tax = 6% of price

To Find:

Cost with tax = ?

Solution:

Cost with tax = Price of laptop + Value of tax

We have to find the value of tax

Value of tax = 6% of Price

putting the value

Value of tax = 6 % * $300

$=\frac{6}{100}*300$

= $18

Now Cost with tax = Price of laptop + Value of tax

Putting in the value

Cost with tax = 300 + 18

= $318

So Cost Including tax would be $318

6 0

3 years ago

Read 2 more answers

What are compatible numbers?

castortr0y [4]

Compatible numbers are numbers that are close in value to the actual numbers and easy to add, subtract, multiply, or divide mentally. They are useful in estimating the sum, difference, product, or quotient.

7 0

3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

Plz help quick. This question is on a quiz in Edg 2020 whoever answered first will get brainliest and I will be giving 50 points

lys-0071 [83]

Answer:

A B and D

Step-by-step explanation:

4 0

2 years ago

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19.

dmitriy555 [2]

Answer:

293 feet per second

Step-by-step explanation:

1056000 feet per hour

1 hour = 60 minutes

1056000 divided by 60 = 17600 feet per minute

1 minute = 60 seconds

17600 divided by 60 = 293 feet per second

5 0

3 years ago