Bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,
![\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2} \\\\\\ sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm\sqrt{1} \\\\\\ sin(2x)=\pm 1\implies sin^{-1}[sin(2x)]=sin^{-1}(\pm 1)](https://tex.z-dn.net/?f=%5Cbf%202sin%5E2%282x%29%3D2%5Cimplies%20sin%5E2%282x%29%3D%5Ccfrac%7B2%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%5E2%282x%29%3D1%5Cimplies%20%5Bsin%282x%29%5D%5E2%3D1%5Cimplies%20sin%282x%29%3D%5Cpm%5Csqrt%7B1%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%282x%29%3D%5Cpm%201%5Cimplies%20sin%5E%7B-1%7D%5Bsin%282x%29%5D%3Dsin%5E%7B-1%7D%28%5Cpm%201%29)
Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
Answer:
Taco supreme is 1230 calories
Pan pizza is 1310 calories
Step-by-step explanation:
Let's denote taco supreme as T and pan pizza az P. We have:
T+2*P = 3850 and 2*T + P = 3770
If we add the equations we get 3*T+3*P = 7620
3*(T+P) = 7620
T+P = 2540 => we can use T=2540-P
Back to one of the first equations we get 2540-P+2*P=3850
P=3850-2540= 1310
=>T= 2540-1310= 1230
Answer:
option A is the answer hope you find it helpful
Answer:
vertex and AOS is 0, and i think the stretch is 1
Step-by-step explanation:
