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lana [24]
3 years ago
5

How to determine if the line is parallel , perpendicular, or neither

Mathematics
1 answer:
4vir4ik [10]3 years ago
5 0
It is parallel if the lines are in straight line next to each other, but don't cross.
It is perpendicular if these lines intersect to form right angles. 
Neither if they don' t follow these rules
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Apples are on sale for $1.20 a pound. Logan bought 3/4 of a pound. How much money did he spend on apples
kati45 [8]
What I did was turn ¾ in to a fraction (which is 75%) then turned it into a decimal (.75)
then I multiplied .75 by 1.20 and .9
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Hope I helped :)
4 0
3 years ago
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In the figure below, segment CD is parallel to segment EF and point H bisects segment DE :
Grace [21]

Answer:

See explanation

Step-by-step explanation:

In the figure below, segment CD is parallel to segment EF, DE is a transversal, then angles DIH and HGI are congruent as alternate interior angles when two parallel lines are cut by a transversal.

Consider triangles DIH and EGH. In these triangles,

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Thus,

\triangle DIH\cong \triangle EGH by AAS postulate

4 0
3 years ago
James is building a sandbox with a width of 4.8 feet and a length of 7.7 feet. ABOUT how much wood will James need to buy to con
4vir4ik [10]

Answer:

it is 25 ft

Step-by-step explanation:

you just add 7.7+7.7+4.8+4.8 to get 25

8 0
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Draw an angle with the given name: <JWJ
Thepotemich [5.8K]
Draw two lines that connect at one point. label where they connect w and the other two ends j
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3 years ago
One uranium atom has a mass of 3.95 x 10^-22 grams.
creativ13 [48]
Remember
(ab)/(cd)=(a/c)(b/d)
and
\frac{x^n}{x^m}=x^{n-m}

1kg=1000g=1*10^3g
hmm, I estimate, 4 of them

x=number of attoms
mass of 1 times x=1*10^3
3.95*10^-22 times x=1*10^3
divide both sides by 3.95*10^-22
x=\frac{1*10^3}{3.95*10^{-22}}= (\frac{1}{3.95}  )( \frac{10^{3}}{10^{-22}} )=0.25316*10^{25}=2.5316*10^{24}
a million has 9 zeroes
so 24 zeroes is alot that is a <span>septillion, so there are 2.5316 septillion uranium atoms in 1kg


a. 4
b. underestimate

</span>
5 0
3 years ago
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