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iVinArrow [24]
3 years ago
10

The length of a rectangle is one less than 3

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
7 0

Answer:

l= 10 , w=7

Step-by-step explanation:

If the length of a rectangle is 11 m less than 3 times the width, then we can come up with this equivalency:

 

l = 3w - 11

 

where l stands for length and w stands for width. Now we plug in the equivalency into the equation for the area of a rectangle:

 

A = lw

A = (3w - 11)w

70 = (3w - 11)w

70 = 3w2 - 11w

 

Then solve for w by completing the square. Now, in order to complete the square for this, we need to find a completion that fits 3w2 - 11w +  ?  and completes the base factoring of (√(3)w -  ? )2. The only one that fits the second question mark so that it can work out to the first is (11√3)/6, and when we work it into the first question mark, the first question mark becomes 121/12. Essentially, what we have for completing the square is:

 

70 + 121/12 = 3w2 - 11w + 121/12

80.08333 = (√(3)w - (11√3)/6)(√(3)w - (11√3)/6)

80.08333 = (√(3)w - (11√3)/6)2

8.94893 = √(3)w - (11√3)/6

 

Solving for w, we get:

 

12.124 = √(3)w

w = 7

 

Since we know that the length is 11 m less than 3 times the width, we can plug w into the first equation to find l:

 

l = 3w - 11

l = 3(7) - 11

l = 21 - 11

l = 10

 

And if we want to check, lw = (10)(7) = 70.

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natka813 [3]

Hello AljakeVillena!

\huge \boxed{\mathfrak{Question} \downarrow}

You pull horizontally on a 50-kg crate with a force of 450 n and the friction force on the crate is 250 n. the acceleration of the crate is ?

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

Given,

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__________________

Let's first find the total force/net force acting on the crate. The total force will be :-

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Now, let's find the acceleration of the crate. To find the acceleration we must divide the net force by the mass. So,..

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__________________

So, the acceleration of the crate is <u>4</u><u> </u><u>m/</u><u>s²</u><u>.</u>

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Hope it'll help you!

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