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3 years ago

What are the possible values of x and y for two distinct points, (5,-2) and (x, y), to represent a function?

Mathematics

1 answer:

gayaneshka [121]3 years ago

3 0

The answer could be (6,3) or ( -3,8)

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Operación 1: (6+(4*12)/12-2)+5

Julli [10]

Answer:

Step-by-step explanation:

6+ (4)(12) /12 −2+5

=6+ 48 /12 −2+5

=6+4−2+5

=10−2+5

=8+5

=13

6 0

2 years ago

Which gives the correct values for points A, B, C, and D?

beks73 [17]

Answer:

The correct answer for this problem is the 2nd one because all you have to do is count how far each one is from -1 and -2

Step-by-step explanation:

5 0

3 years ago

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D=C(1+rt/100) solve for r

xz_007 [3.2K]

Answer:

r= −100c+100d/ct

Step-by-step explanation:

d=c(1+rt/100)

1/100crt+c=d

1/100crt+c+−c=d+−c

1/100crt=−c+d

$\frac{1/100crt}{ct/100}$ = $\frac{-c + d}{ct/100}$

4 0

3 years ago

A fair coin is flipped twelve times. What is the probability of the coin landing tails up exactly nine times?

seraphim [82]

Answer:

$P\left(E\right)=\frac{55}{1024}$

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

2¹² = 4096

We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

$nCr=\frac{n!}{r!\left(n-r\right)!}$

Plug in n = 12 and r = 9

$=\frac{12!}{9!\left(12-9\right)!}$

$=\frac{12!}{9!\cdot \:3!}$

$=\frac{12\cdot \:11\cdot \:10}{3!}$ ∵ $\frac{12!}{9!}=12\cdot \:11\cdot \:10$

$=\frac{1320}{6}$ ∵ $3!\:=\:3\times 2\times 1=6$

$=220$

Thus, the probability will be:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$=\frac{220}{4096}$

$=\frac{55}{1024}$

Thus, the probability of the coin landing tails up exactly nine times will be:

$P\left(E\right)=\frac{55}{1024}$

4 0

3 years ago

Oscar is 16 years old and his sister Julia is 12 hears old. What will be the ratio of Oscar's age to Julia's age in 2 years?

Ivan

The ratio will be 18:14.

4 0

3 years ago

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