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Crank
3 years ago
10

3.which lines are parallel 4. what wouldbyou use to prove these lines parallel

Mathematics
1 answer:
uysha [10]3 years ago
6 0
3)the line G & H are parallel
4) Parallel line theorem is what I do believe it's called (do you have choices for this one?)
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Multiple choice
Aliun [14]
N <span>≥  60 because you take Craig's weight out of the 330 pounds which leaves you with 240 pounds. Then you divide that by 4 to see the highest amount of packages Craig could bring with him. </span>
7 0
2 years ago
Read 2 more answers
Can you help me solve this please step by step
eimsori [14]
First, the values of 2sqrt(3) or 3sqrt(3) are not even twice the length of the given base, 3.

Next, I would say that 6 is the better answer than 12, assuming that the triangle is”drawn-to-scale”.

Finally, I think the hypotenuse is 6.

8 0
2 years ago
Good evening, I am taking an Algebra class for the first time after 50 years, and can not remember how to find the point-slope f
Olin [163]
m=-1/5p x 10.4
m=-1/5p x 52/5
m=-52/25p

8 0
3 years ago
3/4=j-1/2 solve the equation check your solution
DIA [1.3K]
3/4= j-1/2
first to add 1/2 to both sides to solve for j 
3/4 + 1/2= j
second you find a common denominator in order to add 3/4 and 1/2, the common denominator would be 4, you would have to change 3/4 at all because the denominator is already 4 but in order to make the denominator of 1/2, 4 you would have to multiply both the denominator and numerator by 2 so 
3/4 + 2/4 = 5/4
so 5/4 =j

to check your answer you plug 5/4 in for j 
3/4 = 5/4 -1/2
again you need to find a common denominator between 5/4 and 1/2 which again would be 4, and again you wouldn't change 5/4 but you would multiply both the numerator and the denominator of 1/2 so 
3/4= 5/4-2/4
5-2= 3 and you would keep the 4 so 
5/4 - 2/4 = 3/4

so j =  5/4
5 0
2 years ago
Help, I’m stuck. Thank you very much
ZanzabumX [31]
If you notice, the container is really just two circles, with a diameter of 4 each, and a square with sides of 10.5.

now, if we just get the area of each circle, keeping in mind their radius is half of the diameter, namely r = 2, and get the area of the 10.5x10.5 square, sum them up, that'd be the surface area of the container.

\bf \stackrel{\textit{2 circle's area}}{2(\pi r^2)}~~+~~\stackrel{\textit{area of the square}}{10.5\cdot 10.5}&#10;\\\\\\&#10;\stackrel{\textit{2 circle's area}}{2(\pi 2^2)}~~+~~\stackrel{\textit{area of the square}}{10.5\cdot 10.5}\implies 8\pi + 110.25
5 0
3 years ago
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