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loris [4]
3 years ago
6

in these triangles, side AB is congruent to side DF, and side BC is congruent to side FG. Determine the values of x and y

Mathematics
1 answer:
vlada-n [284]3 years ago
5 0
I added a screenshot with the complete question

Answer:
x = 3
y = 9

Explanation:
1- getting the value of x:
We are given that:
side AB is congruent to side DF. This means that:
AB = DF
3(2x+10) = 12x + 12
6x + 30 = 12x + 12
12x - 6x = 30 - 12
6x = 18
x = 18/6
x = 3

2- getting the value of y:
We are given that:
side BC is congruent to side FG. This means that:
BC = FG
2y + 12 = 2(2y-3)
2y + 12 = 4y - 6
4y - 2y = 12 + 6
2y = 18
y = 18/2
y = 9

Hope this helps :)

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How many base pieces 8 1/4 inches long and 8 1/4 inches wide can Mr. Penny cut from a wooden plank 4 1/8 feet long and 4 1/8 fee
podryga [215]

Answer:

36 pieces

Step-by-step explanation:

4 1/8 ft x 4 1/8 ft = 49.5 in x 49.5 in = 2450.25 in^2

8.25 in x 8.25 in = 68.06 in^2

2450.25/68.06 = 36 pieces

checking for scrap loss:

49.5/8.25 = 6

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3 years ago
Simplify 3x-2+5x+9 pls
vfiekz [6]

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8x+7

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 2
Shalnov [3]

Answer:

The expected winnings for a person buying 1 ticket is -0.2.                  

Step-by-step explanation:

Given : A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 25 cents each, find the expected winnings for a person buying 1 ticket.

To find : What are the expected winnings?    

Solution :

There are one first prize, 2 second prize and 20 third prizes.

Probability of getting first prize is \frac{1}{20000}

Probability of getting second prize is \frac{2}{20000}

Probability of getting third prize is \frac{20}{20000}

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each.

So, The value of prizes is

\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10

If 20000 tickets are sold at 25 cents each i.e. $0.25.

Remaining tickets = 20000-1-2-20=19977

Probability of getting remaining tickets is \frac{19977}{20000}

The expected value is

E=\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10-\frac{19977}{20000}\times 0.25

E=\frac{1000+600+200-4994.25}{20000}

E=\frac{-3194.25}{20000}

E=-0.159

Therefore, The expected winnings for a person buying 1 ticket is -0.2.

3 0
3 years ago
Need help homies, no steps needed
VashaNatasha [74]

Answer:

12 have a good day

Step-by-step explanation:

4 0
3 years ago
Elsa and Anna share $2000 in the ratio 5 : 3. How much does each of them get?
Karolina [17]

Answer:

Let them be "x"

5x+3x= $2000

8x= $2000

x= 2000/8

x= 250

5* 250= 1250.

3*250= 750.

5 0
3 years ago
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