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2 years ago

What is 4 times as much as 20.075? show work

Mathematics

1 answer:

choli [55]2 years ago

5 0

Answer:

80.3

Step-by-step explanation:

20.075 x 4 = 80.3

Hope that helps!

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It takes Skyler 10 hours to rake the front lawn while his brother, Sergio, can rake the lawn in 6 hours. How long will it take

PtichkaEL [24]

The answer is 15/4 , or 3.75 or 3 3/4

6 0

2 years ago

Complete the input output table for the function y=3^x

egoroff_w [7]

Answer:

Input of x when it is equal to (1,2,3,4,5) gave an output of 3, 9, 27,81 & 243 respectively.

Step-by-step explanation:

Step 1. Substitute for x (1,2,3,4,5) in the functions to solve for the output of y.

y=3^x

when x= 1

y=3¹ = 3

when x= 2

y=3² = 9

when x= 3

y=3³ = 27

when x= 4

y=3⁴ = 81

when x= 5

y=3 ^ 5 = 243.

Input of x when it is equal to (1,2,3,4,5) gave an output of 3, 9, 27,81 & 243 respectively.

6 0

3 years ago

What is the area of a regular nonagon with 10 cm sides and an apothem if 6cm?

bearhunter [10]

Answer:

The area of the nonagon is 270cm²

Step-by-step explanation:

The area of a regular polygon is given by

$= \frac{1}{2} ap$

where 'a' is the apothem and 'p' is the perimeter.

The apothem is a=6cm.

The perimeter is p=9×10=90cm.

We substitute the apothem and the perimeter to get:

$= \frac{1}{2} \times 6 \times 90 {cm}^{2}$

$= 270 {cm}^{2}$

8 0

3 years ago

What is the distance between the points (4, 6) and (5, 3)? Write the answer in simplest radical form.

frez [133]

Answer:

$\sqrt{10} \: \: units$

Step-by-step explanation:

$\sqrt{(6 - 3)^{2 } + (4 - 5)^{2} }$

$\sqrt{9 + 1}$

$\sqrt{10}$

= 3.16 units

7 0

2 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