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sveticcg [70]
3 years ago
8

describe how do you know if two lines are parallel if all you knew were the equations of the two lines

Mathematics
2 answers:
Marrrta [24]3 years ago
7 0
Two lines are parallel if they intersect.
Andreyy893 years ago
5 0

Answer:

Step-by-step explanation:

Put what you are given in the form of

y = mx + b

If both equations have the same m, the lines are parallel. For example

y = 4x + 10

y = 4x + 20

These two lines are parallel.

However

y = 8x + 40

y = 4x + 10   are not parallel even though it looks like m could be made the same by taking out 4 in each. Check them out on the graph below.

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Step-by-step explanation:

There’s an all called photo math, download it on ur phone and ur set

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3 years ago
Helppppp plsssss!!!<br><br>Thanks!!
Anvisha [2.4K]

Answer:

D

Step-by-step explanation:

using the rule of exponents

• (a/b)^{n} = a^{n} / b^{n}, hence

(7/4)^{11} = 7^{11} / 4^{11} → D



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4 years ago
15+21 the distributive property to factor out the greatest common factor.
vladimir2022 [97]
They are both divisible by 3 so we divide both by 3 and get
3(5+7)
8 0
3 years ago
You're planning to visit with seven clients today and you estimate that at most each client will require 3/4 of an hour of your
erma4kov [3.2K]

Answer:

45 mins

Step-by-step explanation:

Estimated time it would take to visit with each client = fraction of time required for one client x 60 minutes

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5 0
3 years ago
Find the height and area of the equilateral triangle ​
sergejj [24]

As here we are given with an equilateral triangle, whose side is 3 cm, and we need to find the height and area of the equilateral triangle. So, for this let's recall that, the area of any equilateral triangle with side a is given by ;

  • {\boxed{\bf{Area_{(Equilateral\:\: Triangle)}=\dfrac{\sqrt{3}}{4}a^{2}}}}

So, if we substitute a = 3, in our Formula, it will yield to

{:\implies \quad \sf Area_{(Equilateral\:\:Triangle)}=\dfrac{\sqrt{3}}{4}(3)^{2}}

{:\implies \quad \sf Area_{(Equilateral\:\:Triangle)}=\dfrac{\sqrt{3}}{4}(9)}

{:\implies \quad \boxed{\bf{Area_{(Equilateral\:\:Triangle)}=\dfrac{9\sqrt{3}}{4}\:\:cm^{2}}}}

Now, as we know that, for any triangle we also have a formula that is ;

  • {\boxed{\bf{Area_{(Triangle)}=\dfrac{1}{2}\times Base\times Height}}}

Now, Here as the triangle is equilateral, so it's base will just be same 3, and if we let height be H, so we will be having

{:\implies \quad \sf \dfrac{1}{2}\times 3\times H=\dfrac{9\sqrt{3}}{4}}

{:\implies \quad \sf H=\dfrac{9\sqrt{3}}{4}\times \dfrac23}

{:\implies \quad \boxed{\bf{Height=\dfrac{3\sqrt{3}}{2}\:\:cm}}}

3 0
2 years ago
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