I'm not sure if I get what you mean, because I don't think I remeber what an area model is. But if it literally means making a model to describe the area, you could draw a square and put 84 on the length or the width and the o.24 on the other. Show that length times width = area. And then multiply. But if I am right, that's not what you are looking for, huh?

4 0

3 years ago

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A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form

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Answer:

$\vec{v}= \text{ or } \approx$

Step-by-step explanation:

Component form of a vector is given by $\vec{v}=$ , where $i$ represents change in x-value and $j$ represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector $\vec{v}=$ , the magnitude is $||v||=\sqrt{i^2+j^2$ .

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being $i$ , one leg being $j$ , and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

$\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}$

To find the other leg, $i$ , we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

$\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}$

Verify that $(9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark$

Therefore, the component form of this vector is $\vec{v}=\boxed{}\approx \boxed{}$

6 0

3 years ago