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4 years ago

Which values for x and y make the statement (x + 3)(y - 2) = 0 true?

Mathematics

2 answers:

Alona [7]4 years ago

6 0

The correct equation will be y= -1+x

PLEASE MARK AS BRAINLIEST ANSWER :)

Nataliya [291]4 years ago

4 0

In order to make this equation equal to 0 we need to find values for x and y that, when added or subtracted, makes 0 on the left side.

Let's start with x. -3 seems good. Why? Because when added to the x it equals to 0. Now with the y. 2 is good. Because when subtracted it equals to 0.

(-3 + 3)(2 - 2) = 0

0=0

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Help please find the measurements of angle W

guapka [62]

the sum of all the angles = 180

13x-11 + 8x-9 + 30x-4 =180

51x-24 = 180

51x = 204

x=4

w= 8x-9= 8×4-9

= 2e°

5 0

3 years ago

Read 2 more answers

A person has a bag containing dimes and nickels. There are a total of 120 coins in the bag, and the total value of the coins is

telo118 [61]

Answer:

There are 65 dimes. There are 55 nickels.

Step-by-step explanation:

This question can be solved using a system of equations.

I am going to say that:

x is the number of dimes

y is the number of nickels.

There are a total of 120 coins in the bag

This means that x + y = 120.

The total value of the coins is $9.25.

The dime is worth $0.10 and the nickel is worth $0.05. So

0.1x + 0.05y = 9.25

System:

$x + y = 120$

$0.1x + 0.05y = 9.25$

From the first equation:

$y = 120 - x$

Replacing in the second:

$0.1x + 0.05y = 9.25$

$0.1x + 0.05(120 - x) = 9.25$

$0.1x + 6 - 0.05x = 9.25$

$0.05x = 3.25$

$x = \frac{3.25}{0.05}$

$x = 65$

$y = 120 - x = 120 - 65 = 55$

There are 65 dimes. There are 55 nickels.

7 0

4 years ago

The checkbook balance of Roger Co. Is $982.50. The bank statement reveals a balance of $1,051.82. The bank statement showed inte

Lubov Fominskaja [6]

the balance would be 1,285.32 iam not completely sure

8 0

3 years ago

Read 2 more answers

Suppose 29 cars start at a car race. In how many ways can the top 3 cars finish the race?

Ilia_Sergeevich [38]

any of the 34 cars can finish first

any of the remaining 33 cars can finish second

any of the remaining 32 cars can finish third

Thus, the top 3 finishes can be done in (34)(33)(32) = 35904 ways

8 0

4 years ago

Find the 90% confidence interval for the variance and standard deviation of the ages of seniors at oak park college if a random

san4es73 [151]

Given that the standard deviation, s = 2.3 and that there are 24 students, thus the degree of freedom is 24 - 1 = 23.

For 90% confidence interval, $\alpha =0.1$ and $\frac{ \alpha }{2} =0.05$

From the chi-square table, the value of $\chi^2_{1- \frac{ \alpha }{2} }=\chi^2_{0.95}=35.172$ and the value of $\chi^2_{ \frac{ \alpha }{2} }=\chi^2_{0.05}=13.091$ .

The 90% confidence interval for $\sigma^2$ is given by:

$\frac{(n-1)s^2}{\chi^2_{1- \frac{ \alpha }{2} }} \ \textless \ \sigma^2\ \textless \ \frac{(n-1)s^2}{\chi^2_{ \frac{ \alpha }{2} }} \\ \\ = \frac{(24-1)(2.3)^2}{35.172} \ \textless \ \sigma^2\ \textless \ \frac{(24-1)(2.3)^2}{13.091} \\ \\ = \frac{23(5.29)}{35.172} \ \textless \ \sigma^2\ \textless \ \frac{23(5.29)}{13.091} = \frac{121.67}{35.172} \ \textless \ \sigma^2\ \textless \ \frac{121.67}{13.091} \\ \\ =3.46\ \textless \ \sigma^2\ \textless \ 9.29$

And the 90% confidence interval for $\sigma$ is given by:

$\sqrt{\frac{(n-1)s^2}{\chi^2_{1- \frac{ \alpha }{2} }}} \ \textless \ \sigma\ \textless \ \sqrt{\frac{(n-1)s^2}{\chi^2_{ \frac{ \alpha }{2} }}} \\ \\ = \sqrt{3.46} \ \textless \ \sigma\ \textless \ \sqrt{9.29} =1.86\ \textless \ \sigma\ \textless \ 3.05$

3 0

3 years ago