4 years ago

Determine the area of a rectangle with a 12.2 feet and a higher of 3.6 feet

Mathematics

1 answer:

jonny [76]4 years ago

5 0

Answer:

43.92 ft if 12.2 ft is refering to the length and 3.6ft is refering to the width

Step-by-step explanation:

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Write the vector u as a sum of two orthogonal vectors, one of which is the vector projection of u onto v, projvu u=(-8,8) , v= (

Jobisdone [24]

Answer:

none of the given options is true

Step-by-step explanation:

Given: u=(-8,8) , v= (-1,2)

To find: vectors $w_1,w_2$ such that $u=w_1+w_2$

Solution:

A vector is a quantity that has both magnitude and direction.

$proj_vu=\frac{u\cdot v}{\left | v \right |^2}(v)\\=\frac{(-8,8)\cdot (-1,2)}{(\sqrt{ (-1)^2+2^2})^2}(-1,2)\\=\frac{8+16}{5}(-1,2)\\=\frac{1}{5}(-24,48)\\=(-4.8,9.6)$

Let $w_1=(-4.8,9.6)$

So,

$w_2=u-w_1\\=(-8,8)-(-4.8,9.6)\\=(-8+4.8,8-9.6)\\=(-3.2,-1.6)$

So, u = $(-4.8,9.6)+(-3.2,-1.6)$

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