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Studentka2010 [4]
3 years ago
13

What are the numerical measures of each angle?

Mathematics
1 answer:
mash [69]3 years ago
6 0

Answer:

<1 and <3 equal 29 degrees

<2 ans <4 equal 151 degrees

Step-by-step explanation:

instructions in pic :)

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Onlig
igomit [66]

Answer:

Cost of one candy = 80 L.L.

Step-by-step explanation:

Let "x" represent cost of a candy

Cost of 12 candies = 12x

Cost of a chocolate bar = 1,250 L.L.

Total cost = 2,210 L.L.

The following equation represents the scenario given:

1,250 + 12x = 2,210

Solve for x (cost of a candy)

1,250 + 12x - 1,250 = 2,210 - 1,250 (subtraction property of equality)

12x = 960

Divide both sides by 12

12x/12 = 960/12 (division property of equality)

x = 80

Cost of one candy = x = 80 L.L.

3 0
3 years ago
What is the interquartile range of the data? 53, 62, 67, 71, 83, 94, 102, 105
Vsevolod [243]
The IQR is 38 bc the Q1 is 62 and Q3 is 100. 100-62=38
6 0
4 years ago
I WILL GIVE YOU 3 COOKIES IF YOU CAN ANSWER THIS QUESTION CORRECTLY
riadik2000 [5.3K]

Answer:

251.2

I used 3.14 as pi

Step by step explanation:

SA= 2(pi*r²) + (pi*d)h

SA= 2 (\pi * 4² ) + (pi*8)*6

<u>you could also do this:</u>

SA=2πrh + 2πr²

7 0
4 years ago
Read 2 more answers
David made a class banner out of a large rectangular piece of paper. He cut a triangular piece out of one side. what is the area
VikaD [51]
The height would be the same as the rectangular piece of paper as well as the length so what you would do is you do heights times Lanks divide that by two and you would get the Square inches of the banner hope it helps
8 0
3 years ago
How many positive integers less than 100 have at least one digit that is a 9? (Let T be the set of positive integers less than 1
Ymorist [56]

Answer:

19

Step-by-step explanation:

Let T be the set of positive integers less than 100 with a 9 in the ten's place.

T = {90, 91, 92, 93, 94, 95, 96, 97, 98, 99}

n(T) = 10

Let O be the set of positive integers less than 100 with a 9 in the one's place.

O = {9, 19, 29, 39, 49, 59, 69, 79, 89, 99}

n(O) = 10

The common element is 99.

T\cap O={99}

n(T\cap O)=1

The union of both sets is

n(T\cup O)=n(T)+n(O)+n(T\cap O)

n(T\cup O)=10+10-1

n(T\cup O)=19

Therefore, 19 positive integers are less than 100 those have at least one 9.

6 0
4 years ago
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