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Aleksandr-060686 [28]
3 years ago
6

Pls do so this geometry problem been difficult and need the answer who knows how to do this 100% answer

Mathematics
1 answer:
gulaghasi [49]3 years ago
4 0

Answer:

16

Step-by-step explanation:

Since line CE is equal to 32 and line DB is the bisector (cut into the middle) of line CE then line CF would be half of line CE. Hence the reason that line CF is equal to 16

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What is the coefficient in the algebraic expression 7 xy2 ?
Monica [59]
The answer is 7!

The number In front of the variable is always the coefficient.
5 0
3 years ago
34(5f−3)=38 How do i find the value of f?
Oduvanchick [21]

f=14/17 or 0.82352941

8 0
3 years ago
Read 2 more answers
What is x, the distance between points A and A'? 2. 4 units 4. 8 units 13. 6 units 14. 4 units.
adelina 88 [10]

The distance between points A (3, -5) and A' (2, -3) is 2.4 units.

Given that,

The points A (3, -5) and A' (2, -3).

We have to determine,

The distance between point A and A'.

According  to the question,

The distance between two points is determined by using the distance formula.

\rm Distance \ formula = \sqrt{(x_2-x_2) ^2+ (y_2-y_2)^2

Then,

The distance between points A (3, -5) and A' (2, -3) is,

\rm Distance  = \sqrt{(2-3) ^2+ ((-3)-(-5))^2}\\\\ Distance  = \sqrt{(-1) ^2+ (2)^2}\\\\ Distance  = \sqrt{1+4}\\\\Distance = \sqrt{5}\\\\Distance = 2.4 \ units

Hence, The distance between points A (3, -5) and A' (2, -3) is 2.4 units.

For more details refer to the link given below.

brainly.com/question/8069952

6 0
2 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
Which system of equation should you use to solve this problem
Alex17521 [72]

Last option is the correct answer.

Step-by-step explanation:

Given that

Let x stand for the solution a and y stand for the solution B

So,

According to statement that

30 gallons have to be made

x+y=30

Solution A contains 29 percent bromine => 0.29x

Solution B contains 53 percent  =>0.53y

And the mixture of 30 gallons contains 45% bromine

0.29x+0.53y=30*0.45\\0.29x+0.53y=13.5

Following equations will be used for solving the problem

x+y = 30\\0.29x+0.53y=13.5

Hence,

Last option is the correct answer.

Keywords: Equations, Variables

Learn more about linear equations at:

  • brainly.com/question/10941043
  • brainly.com/question/10978510

#LearnwithBrainly

4 0
3 years ago
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