Which of the following is the equation of a circle that has a radius of 2.5 and its center at (4, -3)?

Use properties to rewrite the given equation. Which equations have the same solution as 3/5x +2/3 + x = 1/2– 1/5x? Check all tha

vodomira [7]

we have

$\frac{3}{5}x+ \frac{2}{3}+x=\frac{1}{2}-\frac{1}{5}x$

Combine like terms in both sides

$(\frac{3}{5}x+ x)+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

we know that

$(\frac{3}{5}x+ x)=(\frac{3}{5}x+ \frac{5}{5}x)=\frac{8}{5}x$

substitute in the expression above

$\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$ -----> equation A

Multiply equation A by $5*3*2=30$ both sides

$30*(\frac{8}{5}x+\frac{2}{3})=30*(\frac{1}{2}-\frac{1}{5}x)$

$48x+20=15-6x$ ---------> equation B

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$48x+6x=15-20$

$54x=-5$ ---------> equation C

Solve for x

$x=-\frac{5}{54} =-0.09$

We are going to proceed to verify each case to determine the solution.

Case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

the case a) is equal to the equation A

so

the case a) have the same solution that the given equation

Case b) $18x+20+30x=15-6x$

Combine like terms in left side

$(18x+30x)+20=15-6x$

$(48x)+20=15-6x$

the case b) is equal to the equation B

so

the case b) have the same solution that the given equation

Case c) $18x+20+x=15-6x$

Combine like terms in left side

$(18x+x)+20=15-6x$

$(19x)+20=15-6x$

$19x+6x=15-20\\25x=-5\\x=-0.20$

$-0.20\neq -0.09$

therefore

the case c) not have the same solution that the given equation

Case d) $24x+30x=-5$

Combine like terms in left side

$54x=-5$

the case d) is equal to the equation C

so

the case d) have the same solution that the given equation

Case e) $12x+30x=-5$

Combine like terms in left side

$42x=-5$

$x=-5/42=-0.12$

$-0.12\neq -0.09$

therefore

the case e) not have the same solution that the given equation

therefore

the answer is

case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

case b) $18x+20+30x=15-6x$

case d) $24x+30x=-5$

7 0

4 years ago

Read 2 more answers

HELP ASAP

katovenus [111]

Answer:

1 algebraic expression

Step-by-step explanation:

2. constants

8 0

3 years ago

There are 20 people at a party 2 are best friends. What is the possibility of them both being pretty?

Oliga [24]

I ain’t never seen two pretty best friends, always one of them gotta be ugly

7 0

3 years ago

A computer retail store has 1414 personal computers in stock. A buyer wants to purchase 33 of them. Unknown to either the retail

Lorico [155]

Answer:

a) 364 ways

b) 45.33% probability that exactly one of the computers will be defective.

c) 54.67% probability that at least one of the computers selected is defective.

Step-by-step explanation:

The computers are chosen without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

14 computers, so N = 14.

3 defective, so k = 3.

3 will be purchesed, so n = 3.

A) In how many different ways can the 3 computers be chosen?

3 from a set of 14. So

$C_{14,3} = \frac{14!}{3!(14-3)!} = 364$

364 ways

B) What is the probability that exactly one of thecomputers will be defective?

This is P(X = 1).

$P(X = 1) = h(1,14,3,3) = \frac{C_{3,1}*C_{11,2}}{C_{14,3}} = 0.4533$

45.33% probability that exactly one of the computers will be defective.

C) What is the probability that at least one of the computers selected is defective?

Either none is, or at least one is. The sum of the probabilities of these events is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = h(0,14,3,3) = \frac{C_{3,0}*C_{11,3}}{C_{14,3}} = 0.4533$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.4533 = 0.5467$

54.67% probability that at least one of the computers selected is defective.

5 0

3 years ago

La medida del vector se llama:

Doss [256]

Step-by-step explanation:

suususisissiaisisjsjx

8 0

3 years ago