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never [62]
3 years ago
13

Someone please help!

Mathematics
1 answer:
NikAS [45]3 years ago
3 0
Im pretty sure its D
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Чу- 8х + 5x - Зу<br> OOoO
natulia [17]
Y-3x

Explanation: do 4y-3y which equals 1 which just equals Y. Then do -8x+5x which equals -3x
3 0
3 years ago
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HELP! My teacher didn't teach half the material! How do I do this!?!?!?!
Sladkaya [172]

Answer:

<h2>x=-0.3</h2>

Step-by-step explanation:

1-(-7.2x+3.5)=5+3(7.4x-1) first parenthesis

1+7.2x-3.5=5+22.2x-3  

7.2x-2.5=2+22.2x (x on one side and numbers on other)

22.2x-7.2x= -2-2.5

15x=4.5 divide by 15

x=-4.5/15

<h2>x=-0.3</h2>

5 0
3 years ago
To draw a line segment perpendicular to XY passing through point x
Andrew [12]
What does perpendicular mean? Well, in elementary geometry it mean 2 lines that intersect at a right angle. (90°).

So for Part one. all you need to do is draw a straight line down from the point X. Name the end point of that line M. 

Now you will connect the line M and Z.

Thats it! I hope this helped!! :)




3 0
3 years ago
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Helppp please i need helppppppppoooo
bixtya [17]
Answer
A)

Explanation


4 0
2 years ago
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WHY CAN'T ANYONE HELP ME? :( Jen Butler has been pricing​ Speed-Pass train fares for a group trip to New York. Three adults and
devlian [24]

Answer:

Children: $13

Adults: $18

Step-by-step explanation:

Well for both sets we can set up the following system of equations,

\left \{ {{3a + 4c = 106} \atop {2a + 3c = 75}} \right.

So first we need to solve for a in the first equation.

3a + 4c = 106

-4c to both sides

3a = -4c + 106

Divide 3 by both sides

<u>a = -4/3c + 35 1/3</u>

Now we plug in that a for a in 2a + 3c = 75.

2(-4/3c + 35 1/3) + 3c = 75

-8/3c + 70 2/3 + 3c = 75

Combine like terms

1/3c + 70 2/3 = 75

-70 2/3 to both sides

1/3c = 4 1/3

Divide 1/3 to both sides

c = 13

Now we can plug in 13 for c in 3a + 4c = 106,

3a + 4(13) = 106

3a + 52 = 106

-52 to both sides

3a = 54

Divide 3 by both sides.

a = 18

<em>Thus,</em>

<em>an adult ticket is $18 and a children's ticket is $13.</em>

<em />

<em>Hope this helps :)</em>

5 0
3 years ago
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