All categories

2 years ago

A and B are vertical angles. If m A = (5.x - 2)° and m B = (3r + 12), then find the value of x

Mathematics

2 answers:

DiKsa [7]2 years ago

8 0

Answer:

galina1969 [7]2 years ago

7 0

Answer:

Step-by-step explanation:

∠A=∠B (V.O.A, Vertically oppt angles are eql)

5x-2=3x+12

5x-3x=12+2

2x=14

∴x=14÷2

x=7°

You might be interested in

What is the greatest number of right angles a triangle can contain?

Firdavs [7]

Answer:

C.0

since it's literally impossible for a triangle to have a right angle only squares and other shapes. triangles are made of acute angles

6 0

2 years ago

7,801 + = 4,890 what is the number

zysi [14]

You subtract the numbers 7,801 and 4890

The answer is 2,911

4 0

2 years ago

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

2 years ago

When using base-10 blocks to divide 452 by 2, how many hundreds flats will be in each equal group? hundreds flats Divide 452 by

daser333 [38]

<h3>Given</h3>

4 hundreds flats; 5 tens rods; 2 ones cubes

The number of hundreds flats in each of 2 equal piles

<h3>Solution</h3>

When 4 flats are divided into two equal groups, each group will have ...

... 2 flats

_____

You can imagine doing this the way a card dealer might: first put 1 flat in each of 2 piles, then do the same for the remaining 2 flats. Each pile will end up with 2 flats.

— — — — —

You will have a problem if you continue with the tens rods. There is an odd number of those, so one of them will have to be exchanged for 10 ones cubes.

8 0

2 years ago

Lines ab and cd are parallel. If MN is 70° then what is my

fgiga [73]

The figure below is a parallelogram which statement must be true? D___________E | | |__________ | F G a.line Df is parallel to line de b. Line Df is parallel to line eg c. Line fg is parallel to line eg d. Line fg is parallel to line df 3. All __ are

3 0

2 years ago