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iVinArrow [24]
2 years ago
15

If LaTeX: m\angle ABF=8s-6m ∠ A B F = 8 s − 6 and LaTeX: m\angle ABE=2\left(s+11\right)m ∠ A B E = 2 ( s + 11 ), find LaTeX: m\a

ngle EBFm ∠ E B F. LaTeX: m\angle EBF=m ∠ E B F = [measure]LaTeX: ^\circ
Mathematics
1 answer:
kolbaska11 [484]2 years ago
8 0

Answer:

<h2><em>2(3s-14)</em></h2>

Step-by-step explanation:

Given the angles ∠ABF=8s-6, ∠ABE = 2(s + 11), we are to find the angle ∠EBF. The following expression is true for the three angles;

∠ABF = ∠ABE + ∠EBF

Substituting the given angles into the equation to get the unknown;

8s-6 = 2(s + 11)+ ∠EBF

open the parenthesis

8s-6 = 2s + 22+ ∠EBF

∠EBF = 8s-6-2s-22

collect the like terms

∠EBF = 8s-2s-22-6

∠EBF = 6s-28

factor out the common multiple

∠EBF = 2(3s-14)

<em></em>

<em>Hence the measure of angle ∠EBF is 2(3s-14)</em>

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What is the numerical sum of the degree measures of ∠DEA and ∠AEB?
inessss [21]
The angles are supplementary, therefore totaling them at 180°.

And because I'm bored, I'll solve for x for you!

(x + 25) + (9x + 10) = 180
10x + 35 = 180
Subtract 35 from both sides.
10x = 145
Divide by 10 on both sides.
x = 14.5
3 0
2 years ago
Read 2 more answers
Bonzo went to a carnival. At the first
Shalnov [3]

Bonzo started with 2.1 dollars

Step-by-step explanation:

Assume that Bonzo has $x

1. Change $x to cents

2. Calculate the remaining money with him after each game

3. Equate the left money in the 3rd game by zero to find x

∵ $1 = 100 cents

∴ $x = 100 x cents

First game

∵ Bonzo has 100 x cents

∵ He paid 10¢ to get in

∴ The money left is (100 x - 10)

∵ He spent half  the money he had left

- That mean the money left with him is the other have

∴ The money left with him = \frac{1}{2} (100 x - 10) = (50 x - 5) cents

∵ He spent 10¢  to get out

∴ The money left after 1st game = (50 x - 5) - 10

∴ The money left after 1st game = (50 x - 15) cents

Second game

∵ Bonzo has (50 x - 15) cents

∵ He paid 10¢ to get in

∴ The money left is (50 x - 15) - 10 = (50 x - 25)

∵ He spent half  the money he had left

∴ The money left with him = \frac{1}{2} (50 x - 25) = (25 x - 12.5) cents

∵ He spent 10¢  to get out

∴ The money left after 2nd game = (25 x - 12.5) - 10

∴ The money left after 2nd game = (25 x - 22.5) cents

Third game

∵ Bonzo has (25 x - 22.5) cents

∵ He paid 10¢ to get in

∴ The money left is (25 x - 22.5) - 10 = (25 x - 32.5)

∵ He spent half  the money he had left

∴ The money left with him = \frac{1}{2} (25 x - 32.5) = (12.5 x - 16.25) cents

∵ He spent 10¢  to get out

∴ The money left after 3rd game = (12.5 x - 16.25) - 10

∴ The money left after 3rd game = (12.5 x - 26.25) cents

∵ He found he had no money left after the 3rd game

∴ Equate the left money with him by zero

∴ 12.5 x - 26.25 = 0

- Add 26.25 to both sides

∴ 12.5 x = 26.25

- Divide both sides by 12.5

∴ x = 2.1

<em>Bonzo started with 2.1 dollars </em>

Learn more:

You can learn more about money in  brainly.com/question/1870710

#LearnwithBrainly

5 0
2 years ago
Rewrite (2x + 8) + 4 using the associative property. Then simplify.
skelet666 [1.2K]

Answer:

2x + 12

Step-by-step explanation:

Example of Associative Property Addition:

a + (b+c) = b + (a+c)

Now, implement it to the given expression:

(2x+8) + 4 = 2x + (8+4)

That simplifies to:

2x + 12

4 0
2 years ago
Read 2 more answers
among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 play
kondaur [170]

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

3 0
2 years ago
Write an inequality that represents the situation: a large box of golf balls has more than 12 balls. Describe how your inequalit
qwelly [4]
12>x

X = the number of golf balls, inequality states that the number of golf balls is over 12
7 0
2 years ago
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