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Tatiana [17]
2 years ago
12

Question 1 (Multiple Choice Worth 1 points)

Mathematics
1 answer:
Bumek [7]2 years ago
4 0

Answer:

The third option: 3(x-4)

Step-by-step explanation:

In this case factoring means we want our x variable to be on its own and not have a coefficient, so:

3x-12

= 3·x + 3·4

= 3(x+4)

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HELP PLS <br> Comparing irrational numbers with calculator<br><br> Compare.
julia-pushkina [17]

Answer:

3 2/3 is greater

Step-by-step explanation:

3 2/3 is 3.6666 while √11 is 3.32

6 0
3 years ago
What is ​ BC ​ ?<br><br><br> Enter your answer in the box.
tangare [24]
Lines AB and AC are equal so you set up an equation like so...
4x + 1 = 2x + 23
Subtract 1 from both sides.
4x = 2x + 22
Subtract 2x from both sides.
2x = 22
Divide.
x = 11
Now you need to plug in 11 into 3x - 8.
3(11) - 8
= 33 - 8
= 25
So BC is 25.
I hope this helps love! :) 
5 0
4 years ago
17. Find the value of the function h(x) = 7x - 11 given that = 8m +3
Naya [18.7K]

Answer:

the answer is d

Step-by-step explanation:

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4 0
3 years ago
Whats answere y+2=-3 (x - 4)​
faust18 [17]

Answer:

 Slope = 6.000/2.000 = 3.000

 x-intercept = 10/3 = 3.33333

 y-intercept = -10/1 = -10.00000

Step-by-step explanation:

Slope is defined as the change in y divided by the change in x. We note that for x=0, the value of y is -10.000 and for x=2.000, the value of y is -4.000. So, for a change of 2.000 in x (The change in x is sometimes referred to as "RUN") we get a change of -4.000 - (-10.000) = 6.000 in y. (The change in y is sometimes referred to as "RISE" and the Slope is m = RISE / RUN)

4 0
3 years ago
Read 2 more answers
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
2 years ago
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