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3 years ago

Which point is a solution to the system;

Mathematics

1 answer:

Monica [59]3 years ago

3 0

Answer:

None of the listed

Step-by-step explanation:

$Solve-for ;x -in\\ 2x+2y =18\\x = 9-y\\Substitute , 9-y -for ,x- in -2x-2y=-6\\-2(9-y)-2y =-6\\18+2y-2y=-6\\18 = -6\\The -statement- is -false \\Answer ; No -Solution$

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On a toy car, 1 inch on the toy represents 5 and 1/3 of a

Olenka [21]

Answer:

0.15

Step-by-step explanation:

I don't know if that is right but sorry if it is wrong but I would check with your teacher or someone for help

6 0

2 years ago

Ricky is filling an 8 inch by 4 inch by 4 1/2 inch rectangular box with packing peanuts. The peanuts are cubes that come in two

RideAnS [48]

First find the volume of the box: 8•4•4.5=144 cubic inches.
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He could do 144 and 0
143 and 8
142 or 16
Or just keep following the pattern... Hope that helped

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2 years ago

HELP WILL MARK BRAINLIEST!

denpristay [2]

Answer:

All the exponents in the algebraic expression must be non-negative integers in order for the algebraic expression to be a polynomial. As a general rule of thumb if an algebraic expression has a radical in it then it isn't a polynomial

so no its not.

3 0

2 years ago

Read 2 more answers

What is the center and radius of the circle with equation (x + 9)2 + (y + 5)2 = 64?

muminat

Hello,

center (-9,-5)
Radius=√64=8

Answer A

3 0

3 years ago

Read 2 more answers

Of interest is to test the hypothesis that the mean length of all face-to-face meetings and the mean length of all Zoom meetings

Goshia [24]

Answer:

Null hypothesis: $\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n$

Alternative hypothesis: $\mu_i \neq \mu_j , i,j =1,2,....,n$

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

Step-by-step explanation:

For this case we wnat to test if all the mean length of all face-to-face meetings and the mean length of all Zoom meetings are the same. So then the system of hypothesis are:

Null hypothesis: $\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n$

Alternative hypothesis: $\mu_i \neq \mu_j , i,j =1,2,....,n$

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

6 0

3 years ago