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VladimirAG [237]
3 years ago
12

What are the coordinates of the point 1/4 of the way from A to B?

Mathematics
1 answer:
lisov135 [29]3 years ago
8 0

Answer:

7 units

Step-by-step explanation:

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22. Solve the system
DaniilM [7]
Equation 1)  x + 6y = 2
Equation 2)  5x + 4y = 36

Multiply all of equation 1 by 5.

1)  5(x + 6y = 2)

Simplify.

1)  5x + 30y = 10
2)  5x + 4y = 36

Subtract equations from one another.

26y = -26

Divide both sides by 26.

y = -1

Plug in -1 for y in the first equation.

x + 6y = 2

x + 6(-1) = 2

Simplify.

x - 6 = 2

Add 6 to both sides.

x = 8

D : (8, -1)

~Hope I helped!~
8 0
3 years ago
Kiera's dance class starts at 4:30 p.m. and ends at 6:15 p.m. How long is her dance class?
qwelly [4]
From 4 PM to 6 PM is 2 hours, and it's 15 minutes after 6, so her dance class is 2 hours and 15 minutes long.
8 0
3 years ago
Read 2 more answers
1. The exponential function modeled below represents the number of square
Umnica [9.8K]

Answer:

600,000 km²

Step-by-step explanation:

Using the exponential regression calculator : the regression fit obtained for the data is :

36510 * (1.08^x)

After 40 years ;

x = 40

Hence,

y = 36510 * 1.08^40

y = 36510 * 21.72452

y = 793,162.27 km

Hence, y is closest to 800,000 km

7 0
3 years ago
PHOTO HERE^<br> help????????
Dmitrij [34]
Please put a thanks so i know your question was answered. It helps since im new here and really boosts my morale :D

if x =4 and y =-3

F = \frac{ x^{2}-4y }{2}  =  \frac{ (4)^{2}-4(-3) }{2}  =  \frac{ 16+12 }{2}  = 14


3 0
3 years ago
The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand
NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in  foreign language=0.20

The probability than an American CEO can not transact business in foreign language=1-0.20=0.80

Total number of American CEOs  chosen=12

a. The probability that none can transact business in a foreign language=12C_0(0.20)^0(0.80)^{12}

Using binomial theorem nC_r(1-p)^{n-r}p^r

The probability that none can transact business in a foreign language=\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}

b.The probability that at least two can transact business in a foreign language=1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))

c.The probability that all 12 can transact business in a foreign language=12C_{12}(0.8)^0(0.2)^{12}

The probability that all 12 can transact business in a foreign language=\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}

5 0
3 years ago
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