All categories

2 years ago

If mZOKL = 80°, then what is mZHKJ? С E А4 N H K M

Mathematics

1 answer:

Nookie1986 [14]2 years ago

5 0

MOKL and mHKJ are vertical angles, so they’re congruent
so mHKJ is also 80°

You might be interested in

The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = (x – 12). What is the stand

Mrrafil [7]

Y= x-11 bc...
y-1=x-12
+1 +1
y=x-11

5 0

2 years ago

Determine whether the ordered pairs (1,4) and 6,

devlian [24]

Answer:

Not solutions.

Step-by-step explanation:

-5x+6y=-5

6y=-5+5x

6y=5x-5

y=5/6x-5/6

-5/6 is about -0.833

And the slope is 5/6.

So for every x there is 5/6y.

This means (1, 4) won't work cause it's too low.

Also 5/6*6=5. this is too low too.

So they aren't solutions.

3 0

1 year ago

Pentru a culege nectar 45 de albine au vizitat, intr-o singura exeditie, unele 60 flori, iar altele 75 de flori . Stiing=d ca au

labwork [276]

69.3 I think hope that helps

7 0

2 years ago

Which of these triangles are translations of Triangle A?

kondaur [170]

Answer:

A B C D E F are all the same shape

8 0

3 years ago

Read 2 more answers

Semicircles

Studentka2010 [4]

Answer:

The area of the shaded portion of the figure is $9.1\ cm^2$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The shaded area is equal to the area of the square less the area not shaded.

There are 4 "not shaded" regions.

step 1

Find the area of square ABCD

The area of square is equal to

$A=b^2$

where

b is the length side of the square

we have

$b=4\ cm$

substitute

$A=4^2=16\ cm^2$

step 2

We can find the area of 2 "not shaded" regions by calculating the area of the square less two semi-circles (one circle):

The area of circle is equal to

$A=\pi r^{2}$

The diameter of the circle is equal to the length side of the square

$r=\frac{b}{2}=\frac{4}{2}=2\ cm$ ---> radius is half the diameter

substitute

$A=\pi (2)^{2}$

$A=4\pi\ cm^2$

Therefore, the area of 2 "not-shaded" regions is:

$A=(16-4\pi) \ cm^2$

and the area of 4 "not-shaded" regions is:

$A=2(16-4\pi)=(32-8\pi)\ cm^2$

step 3

Find the area of the shaded region

Remember that the area of the shaded region is the area of the square less 4 "not shaded" regions:

$A=16-(32-8\pi)=(8\pi-16)\ cm^2$

---> exact value

assume

$\pi =3.14$

substitute

$A=(8(3.14)-16)=9.1\ cm^2$

8 0

3 years ago