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)
1)
x^2 + (3y/2z) = 7
2x^2z + 3y = 14z
3y = 14z - 2x^2z
3y = 2z(7 - x^2)
y = 2/3(z)(7 - x^2)
2)
(3zx^4) /(5+z) = 2y
3zx^4 = 2y(5+z)
3zx^4 = 10y + 2yz
Answer:
The first row
Step-by-step explanation:
Letters A,B,C,D
Observe the sequences below. I. 3, 6, 9, 12, ... II. 3, 9, 27, 81, III. 2, 4, 8, 16, ... IV. 3, 5, 7, 9, Which of these are geom
Sholpan [36]
Observe the sequences below. I. 3, 6, 9, 12, ... II. 3, 9, 27, 81, III. 2, 4, 8, 16, ... IV. 3, 5, 7, 9, Which of these are geometric sequences? III only O Il and me II and IV O I and
The dimensions that would result to maximum area will be found as follows:
let the length be x, the width will be 32-x
thus the area will be given by:
P(x)=x(32-x)=32x-x²
At maximum area:
dP'(x)=0
from the expression:
P'(x)=32-2x=0
solving for x
32=2x
x=16 inches
thus the dimensions that will result in maximum are is length=16 inches and width=16 inches
