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Nata [24]
3 years ago
5

Help anyone? I need the questions worked out it doesn’t matter if you give me only one question

Mathematics
1 answer:
Nonamiya [84]3 years ago
7 0

y2-y1/x2-x1. For A and B y2=4, y1=-4, x2=2 and x1=-2... Now u plug them in them, 4-(-4)/2-(-2)= 8/4=2. Slope equals 2. :)

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Write and solve an inequality that represents the values of x, in feet.
SVEN [57.7K]

Answer:

x < 16 ft

Step-by-step explanation:

Perimeter = x

If it is less than 16, the equation is:

x < 16

-Chetan K

3 0
3 years ago
PLS HELPP ITS URGENT
WITCHER [35]

Answer:

Step-by-step explanation:

The angles between the parallel sides add up to 180 degrees.

So 180-70 = 110 degrees.

8 0
3 years ago
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What is the simplified form of the square root of 80?
djyliett [7]
\sqrt{80} =  \sqrt{2 * 2* 2* 2* 5}  =  \sqrt{2^{2}*2^{2}*5}  = 2*2* \sqrt{5} = 4 \sqrt{5}
Hope this helps.
3 0
4 years ago
Which number should each side of the equation 4/5x=8 be multiplied by to produce he equivalent equation of x=10?
gulaghasi [49]
<< Add .\frac{-4}{5} to the left and right of the equation, multiple and make ''opposite multiplation'' to find x.

A) \frac{4}{5x}. \frac{-4}{5} = 8.  \frac{-4}{5} &#10;
\frac{16}{25x} = \frac{-32}{5}
25x.(-32)= (-16).5
-800x= -80

**to leave x alone and positive, you must divide to -800.
x=  \frac{-80}{-800}
x=  \frac{1}{10}
= INCORRECT

B) \frac{4}{5x} . \frac{1}{5} = 8. \frac{1}{5} &#10;
\frac{4}{25x} = \frac{8}{5}
25x.8= 4.5
200x=20&#10;
x=  \frac{20}{200} =  \frac{1}{10}


C)\frac{4}{5x} . \frac{5}{4}=  \frac{8.5}{4}
\frac{1}{x} = \frac{40}{4}
\frac{1}{x} =10
10x=1
x= \frac{1}{10}


D) \frac{4.5}{5x}  = 8.5
\frac{4}{x} = 40
40x= 4
x= \frac{4}{40} = 1/10

I think there is a problem with the options :D







5 0
4 years ago
Which is the solution to the equation 0.5 x + 4.2 = 5.9? Round to the nearest tenth if necessary. 0.9 3.4 5.1 20.2
Anarel [89]

Answer:

x = 3.4

Step-by-step explanation:

Step 1: Isolate <em>x</em>

0.5x = 1.7

Step 2: Divide both sides by 0.5

x = 3.4

And we have our final answer!

8 0
3 years ago
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