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FinnZ [79.3K]
3 years ago
9

Helppppp meeee pleassseee

Mathematics
1 answer:
Elden [556K]3 years ago
5 0
D because it doubles pretend u push the legs together
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HELP ME PLEASE WITH 1,2 AND 3 PLEASE DON'T TAKE LONG TO RESPOND !!​
soldi70 [24.7K]

Answer:

1. -19x

3. 23 -  8x

5. -18x + 2y

Step-by-step explanation:

first: 3x - x = 2x

2x - 22x = -20x

-20x + x = -19x

second one:

8 - (-15) = 23

12x - 20x = -8 x

23 - 8x

third:

-10x - 8x = -18x

3y - y = 2y

-18x + 2y

6 0
3 years ago
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Help please!!! ty ;D
solmaris [256]

Answer:

Jane is 52 7/40 inches tall

Step-by-step explanation:

What you do is you subtract 1 3/8 from 54 3/4 which gives you Carl's height, which is 53 3/8.

Lastly, you subtract 1 1/5 from 53 3/8 which is 52 7/40.

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3 years ago
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What does evaluate each expression mean in <br> seventh grade math
USPshnik [31]
Evaluating each expression means to simplify the expression down to it's simplest form.
7 0
4 years ago
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The graph shows f(x) = (one-half) Superscript x and its translation, g(x). Which describes the translation of f(x) to g(x)?
sveta [45]
There is no graph.
The statement that describes the translation is that it does not exist.
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4 years ago
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A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.
kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is  x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is  −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is  y=3x−13  

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is  13 , the slope of a line perpendicular to line BC is  −3 . The equation of altitude 3 is  y=−3x−1  

The orthocenter is the point where all three altitudes intersect.

x=2  

y=3x−13  

y=−3x−1  

Use substitution to solve the first two equations  y=3(2)−13=−7  

The orthocenter is the point  (2,−7)  

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1  

−7=−6−1  

−7=−7  it works.

3 0
3 years ago
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