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bekas [8.4K]
3 years ago
15

The width of a rectangle is 20 units.Can the perimeter be 64 units when it’s length x is 11

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0

The formula for perimeter is 2L +2W

Replace w with 20 and L with 11:

2(11) + 2(20) = 22 + 40 = 62

This does not equal 64, so the perimeter cannot be 64 with 20 and 11 being the dimensions.

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Find the angle between vector u=2i+sqrt(11)j and v=-3i-2j to the nearest degree
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Answer: Last option 155°

Step-by-step explanation:

We have the components of the vectors u and v.

Then, to find the angle between them, perform the following steps:

1) Calculate the scalar product u * v

If u = 2i + \sqrt{11}j  and v = -3i-2j

Then, the product scalar u * v is:

u * v = (2)(-3) + (\sqrt{11})(-2)

u * v = -6-2\sqrt{11}

u * v = -12.633

2) Calculation of the magnitude of both vectors.

| u | = \sqrt{(2) ^ 2 + (\sqrt{11})^2}\\\\| u | = \sqrt{4+11}\\\\| u | = \sqrt{15}

| v | = \sqrt{(- 3) ^ 2 + (- 2) ^ 2}\\\\| v | = \sqrt{9 +4}\\\\| v | = \sqrt{13}

3) Now that you know the product point between the two vectors and the magnitude of each, then use the following formula to find an angle

u * v = | u || v |cos(\alpha)

-12.633 = \sqrt{15}\sqrt{13}*cos(\alpha)\\\\cos(\alpha) = \frac{-12.633}{\sqrt{15}\sqrt{13}}\\\\arcos(\frac{-12.633}{\sqrt{15}\sqrt{13}}) = \alpha\\\\\alpha =155\°

7 0
4 years ago
I want a correct answer you can take your time. If I was born on December 24, two thousand and four and my classmate was born on
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ASHA 777 [7]

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