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Fantom [35]
3 years ago
15

What is AB and C awnser

Mathematics
2 answers:
hoa [83]3 years ago
8 0

Answer:

20 i think but look at others

Step-by-step explanation:

mr_godi [17]3 years ago
3 0
A and C=58°
B= 58°

You can tell that the triangle is isosceles because of the two congruent sides. In an isosceles triangle, when two sides are congruent, their corresponding angles also are. So by that, you can determine A=C. You then set up the equations equal to each other.

5x+16 = 2x +43
3x+16 = 43
3x = 37
x=9

Then, plug x in.

5(9) + 16 = 61

and

2(9)+43 = 61

After that, you can add up 61 and 61, which equals 122. All triangles add up to 180, so you would do 180-122.

180-122 = 58, so A and C = 61 and C= 58.
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How do you write 100, in an exponential form
Allushta [10]

1. 10^2

2. 10^5

3. 10^4

Hope this helped!

Nate

8 0
3 years ago
What is the coordinates of {2, 1} and {2, 4}?
blondinia [14]
Xm=(2+2)/2=2 Ym=(1+4)/2=2.5 (2,2.5)
7 0
3 years ago
I used 2/3 for tickits and 1/6 on popcorn how much money is left
posledela

Answer:

You would have 1/6 of your money remaining.

Step-by-step explanation:

Let's write an equation for this.

1 - (2/3 + 1/6)

First, let's multiply the numerator and denominator of 2/3. You get 4/6!

1 - (2/3 + 1/6)

1 - (4/6 + 1/6)

Next, let's add 4/6 and 1/6. You get 5/6!

1 - (4/6 + 1/6)

1 - 5/6

Finally, let's turn 1 into 6/6 and do 6/6 - 5/6!

1 - 5/6

6/6 - 5/6

1/6

You would have 1/6 of your money remaining.

3 0
3 years ago
Write a word problem that can be solved by finding the numbers that have 4 as a factor
Ksenya-84 [330]
8 10 12 14 16 I think
5 0
3 years ago
Two numbers have an average of 18, and a difference of 10. Form two equations and solve them to find these
Troyanec [42]
Answer: x = 23 and y = 13

Explanation:

let x and y be the two number.

Where:

(x + y)/2 = 18 => x + y = 18 x 2 = 36

and:

x - y = 10

Thus we get:

x + y = 36 (1)
x - y = 10 (2)

Add (1) and (2):
x + y + x - y = 36 + 10
2x = 46
x = 23

If x = 23

then x - y = 10
23 - y = 10
-y = -13
y = 13
7 0
3 years ago
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