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Papessa [141]
3 years ago
14

WHICH NUMBERS ARE SOLUTIONS OF EACH INEQUALITY?:

Mathematics
1 answer:
love history [14]3 years ago
4 0

Answer:

I don't know if i am right but i think it is number 7 because it like goes in order

Step-by-step explanation:

i am so unsure so i think it is 7 i don't know for sure

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Really need help not good in geometry
Mandarinka [93]

Answer:18

Step-by-step explanation:It is the same shape just 3 times the size.

4 0
3 years ago
F(x) = 4x - 9 <br> i need helpp
Masteriza [31]

Answer:

Step-by-step explanation:

10

5 0
3 years ago
Read 2 more answers
Pleaseee helppppppppppppppppppp
quester [9]

To find Angle A we use cosine

cos ∅ = adjacent / hypotenuse

From the question

The adjacent is 17

The hypotenuse is 38

So we have

cos A = 17/38

A = cos-¹ 17/38

A = 63.4

<h3>A = 63° to the nearest degree</h3>

To find Angle C we use sine

sin ∅ = opposite / hypotenuse

From the question

The opposite is 17

The hypotenuse is 38

So we have

sin C = 17/38

C = sin-¹ 17/38

C = 26.57

<h3>C = 27° to the nearest degree</h3>

Hope this helps you

7 0
3 years ago
Help!!!! please!!!! ill give brainliest!!!!
Anna [14]

Answer: The 2nd option is correct.

Step-by-step explanation:

Since you have started off with 4 players, the number line needs to start at 5, because 9 - 4 = 5. This gives us the answer of the second option, because it starts at 5 and goes to the positive side.

3 0
3 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
2 years ago
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