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3 years ago

Area of a 40cm semicircle

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

3 0

20 i think sorry if its wrong not my best subject.

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A cylinder with a height of 12 cm is inscribed in a sphere with a radius of 10 cm. Find the lateral area of the cylinder. Answer

zalisa [80]

Answer:

256π

Step-by-step explanation:

Given that:

The height of the cylinder: 12 cm

The radius of the sphere : 10

Let r is the radius of the cylinder, use Pytagon

10² = r² + 6²

<=> r² = 10² - 6² = 64

<=> r = 8

Hence, the lateral area of the cylinder

L= 2πrh

= 2π8*16

= 256π

4 0

3 years ago

Anybody body willing to help i have to do 1-3 and 14-17 for all of my hw

Ymorist [56]

I will help, what is your book called?

6 0

3 years ago

Noah and Lin are each trying to solve the equation x² - 6x + 10 = 0. They know that the solutions to x² = - 1 are i and -i, but

Leviafan [203]

Answer:

Solving the equation $x^2 - 6x + 10 = 0$ we get $x=3+i \ , \ x=3-i$

Step-by-step explanation:

Noah's step to solve the equation $x^2 - 6x + 10 = 0$ are:

$x^2 - 6x + 10 = 0\\x^2 - 6x = -10\\x^2 - 6x + 9 = -10 + 9\\(x-3)^2 = -1$

The next step is to take square root on both sides because: $\sqrt{x^2}=x$

$\sqrt{(x-3)^2}=\sqrt{-1} \\x-3=\pm\sqrt{-1} \\x=\pm(\sqrt{-1})+3 \\$

Now, we know that: $\sqrt{-1}=i$

$x=\pm(i)+3\\x=3+i \ , \ x=3-i$

So, Solving the equation $x^2 - 6x + 10 = 0$ we get $x=3+i \ , \ x=3-i$

8 0

3 years ago

Literal equations 5y-72=7+5x

joja [24]

Step 1. Add 72 to both sides

5y = 7 + 5x + 72

Step 2. Simplify 7 + 5x + 72 to 5x + 79

5y = 5x + 79

Step 3. Divide both sides by 5

y = 5x + 79/5

4 0

3 years ago

Based on the information below, what are the values of x and y of the solution to the system of equations used to create the inf

Alborosie

Answer:

Option D) is correct

That is x=2 , y=-5

Step-by-step explanation:

Given that the $|A|=\left| \begin{array}{cc}4&-6\\ 8&-2\end{array}\right|, \left|A_x\right|=\left|\begin{array}{cc}38&-6\\ 28&-2\end{array}\right|, \left|A_y\right|=\left|\begin{array}{cc}4&38\\ 8&26\end{array}\right|$

To find their determinants

$|A|=\left| \begin{array}{cc}4&-6\\ 8&-2\end{array}\right|$

=-8+48

=40

Therefore $|A|=40$

$|A_x|=\left|\begin{array}{cc}38&-6\\ 28&-2\end{array}\right|$

=-76+156

=80

Therefore $|A_x|=80$

$|A_y|=\left|\begin{array}{cc}4&38\\ 8&26\end{array}\right|$

=104-304

=-200

Therefore $|A_y|=-200$

To find the values of x and y:

$x=\frac{|A_x|}{|A|}\ \textrm{and}\ y=\frac{|A_y|}{|A|}$

$x=\frac{80}{40}$ and $y=\frac{-200}{40}$

x=2 and y=-5

Option D) is correct

That is x=2 , y=-5

7 0

3 years ago

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