For this exercise you need to use the formula for calculate the magnitude of the resultant vector obtained by adding the vectors "a" and "b" shown in the picture attached:

$c=\sqrt{a^2+b^2}$

Where "c" is the magnitude of the resultant vector asked in the exercise.

You need to analyze the graph given in the exercise.

You can identify that:

$a=4\\\\b=3$

Then the next step, knowing which are "a" and "b", is to substitute values into the formula given at the beginning of this explanation, as following:

$c=\sqrt{4^2+3^2}$

Finally, evaluating, you get that "c" is:

$c=\sqrt{16+9}\\\\c=5$

4 0

3 years ago

If sin theta = 3/4 and is in the first quadrant, then cos theta

Rashid [163]

Answer:

$\frac{\sqrt{7} }{4}$

Step-by-step explanation:

since sin theta = 3/4, opposite length = 3 units and hypotenuse length = 4 units

by pythagoras' theorem, adjacent length = $\sqrt{7}$

thus cos theta = adjacent/hypotenuse = $\frac{\sqrt{7} }{4}$

7 0

1 year ago

Read 2 more answers

x+4, -x+64. Use the expressions that represent the length and width of the game board to write an equation that models the area

AfilCa [17]

Y=(x+4)(-x+64)
y=-x^2+68

8 0

3 years ago