All categories

3 years ago

What is the approximate area of the shaded region?

Mathematics

2 answers:

vladimir2022 [97]3 years ago

8 0

The answer is 13.76 meters squared.

Alla [95]3 years ago

7 0

Approximately 13.73
The Square's area is 64 and the circle's area is approximately 50.27
the difference is your answer

You might be interested in

Help plzzzz it’s a timed test

ad-work [718]

Answer:

k= -1. 1

Step-by-step explanation:

distribution the 8

32k- 8k=-5k -32 then bring the K to the other side and add .

29k=-32 then divide and you get

k= -1.1 rounded

5 0

3 years ago

WHAT VARIES MORE 6.6 OR 4.3: NO LINKS

Svetllana [295]

Answer:

what exactly do you mean by that?

Step-by-step explanation:

**i promise to go back and provide a real answer** (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧

8 0

2 years ago

Read 2 more answers

Gravity pulls the metal down at 10 N, but the metal pieces RISES! At least how much force is needed to make it move upward?

prisoha [69]

Answer: Greater than 10 N

Step-by-step explanation:

Given

Gravity force on metal is 10 N

But the metal piece is rising i.e. there must be an upward force greater than 10 N

Suppose the force is F

So, we can write

$\Rightarrow F>10\ N$

Any force greater than 10 N is sufficient for the metal to move upwards

The greater the amount of force faster will be the movement

3 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

What is the range of possible sizes for side x?<br> 3.2<br> x<br> 5.5

ladessa [460]

Answer: try multiplication for the answer or use photomath

Step-by-step explanation:

3 0

2 years ago