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irina1246 [14]
3 years ago
11

Angle A and B are congruent angles.•The measure of angle

Mathematics
1 answer:
grigory [225]3 years ago
6 0

Answer:

x = 8

Step-by-step explanation:

8x + 9 = 41 + 4x

4x = 32

x = 8

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25.9 m

Step-by-step explanation:

Add them all together.

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Plz help me well mark brainliest if you are correct!..
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8

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all numbers must be positive when you have the lines on the side

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Plz help I need help
laiz [17]

Answer:

The answer is most likely -1 31/33

I have a hard time with fractions so I usually just go into my calculator and change it to a decimal. For example with -5 1/3 I would insert 1/3 into my calculator to change it to a decimal and then simply tack on the -5 to it. So in decimal form 1/3 is 0.333... so with the -5 added on its -5.333... Do the same for the other number and then divide. You will get -1.9393... all you have to do is go into your calculator and try out 11/12 or 31/33 to see which equals 0.9393... because those are the only options with a -1. In this case 31/33 equals 0.9393 s -1 31/33 is the answer

5 0
2 years ago
What is mean median mode
Alborosie

Answer:

Step-by-step explanation:

mean- add up all numbers and divide by how many numbers there are

median- the middle number out of the list of numbers

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5 0
3 years ago
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Tacoma's population in 2000 was about 200 thousand, and has been growing by about 8% each year. If this continues, what will Tac
Anestetic [448]

Answer:

Tacoma's population in 2014 will be 587,439 people

Step-by-step explanation:

we know that

The equation of a exponential growth function  is given by the formula

y=a(1+r)^x

where

y is Tacoma's population in thousand

x is the number of years since 2000

r is the rate of change

a is the initial value

we have

a=200\\r=8\%=8/100=0.08

substitute

y=200(1+0.08)^x

y=200(1.08)^x

what will Tacoma's population be in 2014?

Find the value of x

x=2014-2000=14\ years

substitute the value of x

y=200(1.08)^{14}=587.439\ thousand

therefore

Tacoma's population in 2014 will be 587,439 people

5 0
3 years ago
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