All categories

3 years ago

What are the adjacent and vertical pairs in this picture

Mathematics

1 answer:

luda_lava [24]3 years ago

5 0

Answer:

Step-by-step explanation:

Two line segments AD and CE intersects at a point B.

Two angles which represent the pair of adjacent angles are,

∠ABC and ∠DBC

Pair of vertical angles (Opposite angles between two intersecting lines) is,

∠ABE and ∠CBD

You might be interested in

Please help me I really need to turn it in I hate khan academy

kow [346]

42.5 units sq
I’m not sure because I did this partly in my head but this should be right.

5 0

3 years ago

What type of number is<br> -4 pi

kobusy [5.1K]

-4 pi is an irrational number

5 0

3 years ago

The midpoints of an irregular quadrilateral ABCD are connected to form another quadrilateral inside ABCD. Complete the explanati

dem82 [27]

Answer:

Suppose: M, N, P, Q are the midpoints of AB, BC, CD, AD respectively

=> MNPQ is the quadrilateral inside ABCD

connect B to D, ΔABD has : M is the midpoint of AB

Q is the midpoint of AD

=> MQ is the midpoint polygon of ΔABD

=> MQ // BD and MQ = 1/2.BD (1)

ΔBCD has: N is the midpoint of BC

P is the midpoint of DC

=> NP is the midpoint polygon of ΔBCD

=> NP // BD and NP = 1/2.BD (2)

from (1) and (2) => MQ // NP ( //BD)

MQ = NP (=1/2.BD)

=> MNPQ is a parallelogram.

=> the quadrilateral inside ABCD is a parallelogram.

Step-by-step explanation:

4 0

3 years ago

Find C. to form completing Square<br> <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-%5Cfrac%7B3%7D%7B2%7Dx%20%2B%20C" id="Tex

nata0808 [166]

Answer:

c = 9/16

Step-by-step explanation:

(a-b)²

a² - 2ab + b²

here a = x

b = ?

x² - (2)(x)(?) + ?²

Now we can see that -(2x)? = -3x/2

-2x? = -3x/2

cancel -x on both sides

2? = 3/2

? = 3/2 * (2)

? = 3/4

so b = 3/4

(x² - 3x + 9/16) = (x-3/4)²

so from the expression we can see that C = 9/16

3 0

3 years ago

What is the radius of a circle whose equation is x2 y2 8x – 6y 21 = 0?

andrew11 [14]

we know that

the standard form of the equation of the circle is

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$x^{2} +y^{2} +8x-6y+21=0$

<u>Convert to standard form</u>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+8x) +(y^{2}-6y)=-21$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+8x+16) +(y^{2}-6y+9)=-21+16+9$

$(x^{2}+8x+16) +(y^{2}-6y+9)=4$

Rewrite as perfect squares

$(x+4)^{2} +(y-3)^{2}=4$

$(x+4)^{2} +(y-3)^{2}=2^{2}$

the center of the circle is $(-4,3)$

the radius of the circle is $2\ units$

therefore

<u>the answer is</u>

the radius of the circle is $2\ units$

8 0

3 years ago

Read 2 more answers