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3 years ago

The radius of a circle is 6mm. What is the area of the shaded portion of the circle.

Mathematics

1 answer:

mariarad [96]3 years ago

6 0

Answer:

b) 29π mm²

Step-by-step explanation:

$\frac{360}{360} - \frac{70}{360} = \frac{290}{360} \\ \\\pi \: {r}^{2} \\ = \frac{290}{360} \times \pi\times 6 \times 6 \\ = 29 \: \pi \: {mm}^{2}$

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Restaurants A, B, C, and D lie on a circular lake. Restaurants A and Care near the

DaniilM [7]

Answer:

Step-by-step explanation:

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3 years ago

What is the rate of change.

Finger [1]

Answer:

-1

Step-by-step explanation:

Over 12 on the x axis it goes down 12 on the y axis so the rate of change is -1

4 0

3 years ago

Can someone help me please And please show your work

kondaur [170]

A)1:2
b)21:2
c)6:44 or 3:22
d)2:5

5 0

3 years ago

In the center of town there is a square park with an area of 900 square feet. If Stephen walks from one corner of the park to th

Vedmedyk [2.9K]

The distance walked by Stephen is 42.42 feet

<h3>Solution:</h3>

Given that In the center of town there is a square park with an area of 900 square feet

Stephen walks from one corner of the park to the opposite corner

To find: Distance walked by Stephen

We need to find the distance that the person walk from one corner of the park to the opposite corner.

So, we need to find the Diagonal of the square park

The diagonal of square is given as:

$\text{ Diagonal } = \sqrt{2} \times \text{ side}$

Let us find the length of side of square

Given area of square = 900 square feet

The area of square is given as:

$area = (side)^2\\\\900 = (side)^2$

Taking square root on both sides,

$side = \sqrt{900} \\\\side = 30$

Thus length of each side of square is 30 feet

Therefore, length of diagonal is given as:

$diagonal = \sqrt{2} \times 30\\\\diagonal = 1.414 \times 30 = 42.42$

Hence, the distance that the person walks 42.42 feet

3 0

3 years ago

2x^2+3x-2, a=0 How do I find the derivative?

Ghella [55]

You can use the definition:

$\displaystyle f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}h$

Then if

$f(x) = 2x^2+3x-2$

we have

$f(x+h) = 2(x+h)^2+3(x+h) - 2 = 2x^2 + 4xh+2h^2+3x+3h-2$

Then the derivative is

$\displaystyle f'(x) = \lim_{h\to0}\frac{(2x^2+4xh+2h^2+3x+3h-2)-(2x^2+3x-2)}h \\\\ f'(x) = \lim_{h\to0}\frac{4xh+2h^2+3h}h \\\\ f'(x) = \lim_{h\to0}(4x+2h+3) = \boxed{4x+3}$

I'm guessing the second part of the question asks you to find the tangent line to f(x) at the point a = 0. The slope of the tangent line to this point is

$f'(0) = 4(0) + 3 = 3$

and when a = 0, we have f(a) = f (0) = -2, so the graph of f(x) passes through the point (0, -2).

Use the point-slope formula to get the equation of the tangent line:

y - (-2) = 3 (x - 0)

y + 2 = 3x

y = 3x - 2

5 0

3 years ago