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2 years ago

Find the magnitude of the

Mathematics

1 answer:

sertanlavr [38]2 years ago

7 0

17.49 is the magnitude of the resultant vector. This can be obtained by adding the vectors and finding magnitude.

<h3>Calculate the magnitude of the resultant vector:</h3>

Resultant vector: vector that gives the combined effect of all the vectors. When we add two or more vectors, the outcome is the resultant vector.

For example,

Sum of vector (3,4) and vector (5,7)

(3,4) + (5,7) = (3+5,4+7) = (8,11)

(8,11) is the resultant vector of vector (3,4) and vector (5,7)

Magnitude of a vector: the length of the vector. The magnitude of the vector a is denoted as ∥a∥.

Magnitude of a = (a₁, a₂)

∥a∥= $\sqrt{a_{1}^{2} +a_{2}^{2} }$

In the given question,

v (8,10) and w (7, -1)

Adding the vectors,

resultant vector = (8,10) + (7, -1) = (15, 9)

Magnitude of a vector (a,b) = √a²+b²

Here, magnitude = √15²+9² = √225+81 = √306 ≈ 17.49

Hence 17.49 is the magnitude of the resultant vector.

Learn more about vectors:

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The volume of this cylinder is 21,352 cubic feet. What is the height? Use a 3.14 and round your answer to the nearest hundredth.

slava [35]

The result is cubit feet 444 ft

8 0

3 years ago

lilavasa [31]

The count of the equilateral triangle is an illustration of areas

There are 150 small equilateral triangles in the regular hexagon

<h3>How to determine the number of equilateral triangle </h3>

The side length of the hexagon is given as:

L = 5

The area of the hexagon is calculated as:

$A = \frac{3\sqrt 3}{2}L^2$

This gives

$A = \frac{3\sqrt 3}{2}* 5^2$

$A = \frac{75\sqrt 3}{2}$

The side length of the equilateral triangle is

l = 1

The area of the triangle is calculated as:

$a = \frac{\sqrt 3}{4}l^2$

So, we have:

$a = \frac{\sqrt 3}{4}*1^2$

$a = \frac{\sqrt 3}{4}$

The number of equilateral triangles in the regular hexagon is then calculated as:

$n = \frac Aa$

This gives

$n = \frac{75\sqrt 3}{2} \div \frac{\sqrt 3}4$

So, we have:

$n = \frac{75}{2} \div \frac{1}4$

Rewrite as:

$n = \frac{75}{2} *\frac{4}1$

$n = 150$

Hence, there are 150 small equilateral triangles in the regular hexagon