All categories

2 years ago

Please help me this is urgent

Mathematics

1 answer:

ahrayia [7]2 years ago

7 0

Answer:

we know,

area of circle = πr²

area of semicircle of circle having diameter of 16.

according to formula = πr²

radius = 16/2 = 8 cm

22/7 × (8)² → 22/7 × 64 → 1408/7 = 201.14/2 = 100.57 cm (approx.)

area of semicircle of circle having diameter of 20 cm.

22/7 × 20² = 22/7 × 400 = 8800/7 = 1257.14/2 = 628.57

to get the area of shaded part 628.57 - 100.57 = 528 cm² (approx.)

hope this helps you!

You might be interested in

How to graph 2x+5y=10

kotegsom [21]

If you solve the equation to get the equation y=mx+b then 2x+5y=10 equals to
y=(-2x-10)/5 . If you use a graphing calculator and type that into y= it should graph it for you going through coordinates (-3,4) , (0,-2) , (3,-8) and so forth

3 0

3 years ago

A diver takes a running leap off of a 40 foot cliff into a pool of water. When she is 5 feet from the cliff,

padilas [110]

of" (and any subsequent words) was ignored because we limit queries to 32 words.

8 0

2 years ago

Can you help with #13 and #14<br><br> sorry about the last one :/

Dima020 [189]

Number 13 is 40 in
number 14 is 2.5

8 0

2 years ago

Read 2 more answers

Find the midpoint of the line segment with the endpoints A and B.

zavuch27 [327]

Answer:

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

$M = ( \frac{x1 + x2}{2} , \: \frac{y1 + y2}{2} )\\$

From the question we have

$m = ( \frac{2 + 10}{2} \: , \: \frac{6 + 8}{2} ) \\ = (\frac{12}{2} \: , \: \frac{14}{2} ) \: \: \: \: \: \: \:$

We have the final answer as

Hope this helps you

3 0

2 years ago

What is the oblique asymptote of the function f(x) x^2+x-2/x+1

Mamont248 [21]

$f(x)=\displaystyle\frac{x^2+x-2}{x+1}=\frac{x(x+1)-2}{x+1}=x-\frac{2}{x+1}
$

$\displaystyle\lim_{x\to\infty}\left\{f(x)-x \right\}=\lim_{x\to\infty}\frac{2}{x+1}=\lim_{x\to\infty}\frac{\displaystyle\frac{2}{x}}{1+\displaystyle\frac{1}{x}}
=0$

Consequently, t<span>he limit of $f(x)$ as x approaches infinity is $x$ .

In other words, $f(x)$ approaches the line y=x,
</span><span>
so oblique asymptote is y=x.

I'm Japanese, if you find some mistakes in my English, please let me know.</span>

3 0

3 years ago