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Elan Coil [88]
3 years ago
7

I do need to know NOTHING.

Mathematics
1 answer:
LuckyWell [14K]3 years ago
8 0

Answer:

Thats cool man.

Step-by-step explanation:

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Write an equation to represent :P
SVEN [57.7K]
$55+$40=95 for 1 hr

95x2=$190

so she came for 2 hrs
7 0
3 years ago
Please help will give brainliest and 80 points
belka [17]

Answer:

Start

A2

B2

B1

C1

C2

D2

D3

D4

C4

END

Step-by-step explanation:

Start (A3)

x is equal to 141 because they are alternate interior angles.

A2. x is equal to 39 because they are corresponding angles.

B2. x would be supplementary to 41 because the angle that x supplements is corresponding to 41.

41 + x = 180 due to the linear pair postulate. Therefore, x = 139.

B1. x would be supplementary to 82 because they are consecutive exterior angles.

82 + x = 180 due to the linear pair postulate. Therefore, x = 98.

C1. x = 102 due to the vertical angles theorem.

C2. x would be supplementary to 130 because the angle that x supplements is equal to 130 (Alternate Exterior Angles).

130 + x = 180, x = 50.

D2. x = 74, corresponding angles.

D3. x = 83, corresponding angles.

D4. x = 95, corresponding

C4. x is supplementary to 18 because of the consecutive interior angles theorem.

x = 162

END

8 0
4 years ago
Read 2 more answers
PLEASE HELP!!!
Brut [27]

Answer:PLEASE HELP!!!

7 0
3 years ago
How do I solve this (-2,6) (4,3) <br>it's a select all correct answers
LuckyWell [14K]
Use a coordinate plane chart to help you
4 0
4 years ago
(5/6)^4x=(36/25)^9-x, please help solve for x, and the 9-x is all superscript
stiks02 [169]
\frac{36}{25}=\frac{6^2}{5^2}=\left(\frac{6}{5}\right)^2=\left(\frac{5}{6}\right)^{-2}\\\\therefore:\\\\\left(\frac{5}{6}\right)^{4x}=\left[\left(\frac{5}{6}\right)^{-2}\right]^{9-x}\\\\\left(\frac{5}{6}\right)^{4x}=\left(\frac{5}{6}\right)^{-2(9-x)}\iff4x=-2(9-x)\\\\4x=-2\cdot9-2\cdot(-x)\\\\4x=-18+2x\ \ \ \ \ |subtract\ 2x\ from\ both\ sides\\\\2x=-18\ \ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{x=-9}


Use:\\a^{-n}=\left(\frac{1}{a}\right)^n\\\\\left(a^n\right)^m=a^{n\cdot m}\\\\\left(\frac{a}{b} \right)^n=\frac{a^n}{b^n}
6 0
3 years ago
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