If tan theta is -1, we know immediately that theta is in either Quadrant II or Q IV. We need to focus on Q IV due to the restrictions on theta.
Because tan theta is -1, the ray representing theta makes a 45 degree angle with the horiz axis, and a 45 degree angle with the negative vert. axis. Thus the hypotenuse, by the Pythagorean Theorem, tells us that the hyp is sqrt(2).
Thus, the cosine of theta is adj / hyp, or +1 / sqrt(2), or [sqrt(2)]/2
The secant of theta is the reciprocal of that, and thus is
2 sqrt(2)
---------- * ------------ = sqrt(2) (answer)
sqrt(2) sqrt(2)
4y ≥ 3x + 2
Plugging in the values in the options, then the required values are when x = 6 and y = 5, then
4(5) ≥ 3(6) + 2
20 ≥ 18 + 2
20 ≥ 20
Answer:
Yes, B
Step-by-step explanation:
Yes the data is linear because there is a constant rate of change.
For x or the input the rate of change is plus 2
For y or the output the rate of change is plus 4
y/x is your slope so 4/2=2
and the only equation with a slope of 2 is B
Answer:
The radius of circle B is 6 times greater than the radius of circle A
The area of circle B is 36 times greater than the area of circle A
Step-by-step explanation:
we have
<em>Circle A</em>
The radius of circle A is
-----> the radius is half the diameter
<em>Circle B</em>
Compare the radius of both circles
The radius of circle B is six times greater than the radius of circle A
Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared
All circles are similar
In this problem the scale factor is 6
so
therefore
The area of circle B is 36 times greater than the area of circle A
