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2 years ago

6 7 8 only 8th grade math

Mathematics

1 answer:

saul85 [17]2 years ago

8 0

☺ I might have misunderstood, but I can help if you give me the question. ☺

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Which of these systems of linear equations has no solution?

Alika [10]

Answer:

A is correct

Step-by-step explanation:

A has no solution because since the slopes are the same, setting both equations equal to each other will cancel out the slopes and result in 8=16, meaning no solution will make both sides equal to each other.

B does have a solution because of different slopes

C does have a solution because of different slopes

D does have a solution because of different slopes

7 0

2 years ago

If (x+3)divided by 3 = 9 what is x

defon

Answer:

x = 24

Step-by-step explanation:

Given that:

(x+3)/3 = 9

multiplying both sides by 3

(x+3) = 9*3

(x+3)= 27

Subtracting 3 from both sides

x= 27-3

x=24

I hope it will help you!

3 0

3 years ago

Read 2 more answers

I need help and i need it fast TIME DUE 4 MIN

Phoenix [80]

Answer:

Step-by-step explanation:

multyply both like 10 x 6 then 11 x 7 after that subtrac the numbers that you múltiple and you will get the answer

8 0

3 years ago

Help me please I'm stuck on this question!!!!

shutvik [7]

Answer:

1: 2.8

Step-by-step explanation:

15:42

Divide each side by 15

15/15 : 42/15

1: 2.8

6 0

2 years ago

Here are the endpoints of the segments BC, FG, and JK.<br> B, −67

yulyashka [42]

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}$

$F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}$

$JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill$

4 0

1 year ago