Answer: Choice B. sqrt(2)
Draw out a right triangle in quadrant IV as you see in the attached image below. The horizontal and vertical legs are both 1 unit long. To ensure that the signs are properly set up, I am making the vertical leg BC have a label "-1" to mean this is below the x axis. Note how
tan(theta) = opposite/adjacent = BC/AB = -1/1 = -1
Use the pythagorean theorem to find that the hypotenuse AC is sqrt(2) units long
a^2 + b^2 = c^2
(1)^2 + (1)^2 = c^2
2 = c^2
c^2 = 2
c = sqrt(2)
The secant of theta is the ratio of the hypotenuse over the adjacent side, so we end up with
sec(theta) = hypotenuse/adjacent
sec(theta) = AC/AB
sec(theta) = sqrt(2)/1
sec(theta) = sqrt(2) which is why choice B is the answer
X = 3/2 or 1 1/2
Both of the equations equal -10, therefor the equations are equal.
Answer:
x = 7, arc DE = 198°
Step-by-step explanation:
Inscribed angles from the same arc are congruent.
∠ DCE and ∠ DWE are inscribed angles from the same arc DE , thus
12x + 15 = 3x + 78 ( subtract 3x from both sides )
9x + 15 = 78 ( subtract 15 from both sides )
9x = 63 ( divide both sides by 9 )
x = 7
Thus
∠ DCE = 12x + 15 = 12(7) + 15 = 84 + 15 = 99°
The inscribed angle DCE is half the measure of its intercepted arc, thus
arc DE = 2 × 99° = 198°
Answer:
The graph of y2+3x=0 is symmetric with respect to the x-axis
Step-by-step explanation:
To establish symmetry with respect to the x-axis we simply substitute -y in place of y in the original equation. If the resulting equation is identical to the original one then the function is said to be symmetric with respect to the x-axis.
In this case we have;
Which is identical to the original equation
The answer is -18
9 · -2 = -<span>18
</span>
