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.
(-2)(-24)
(-3)(-16)
(-4)(-12)
(-6)(-8)
-(-1)(-48)
1*48
2*24
3*16
4*12
6*8
Answer:
x ≈ 38.7°
Step-by-step explanation:
Using the sine ratio in the right triangle
sinx = 
= 
, then
x = 
( ) ≈ 38.7° ( to 1 dec. place )
We have an isosceles triangle;
A=opposite angle side a.
B=opposite angle side b.
C=opposite angle side c.
A=B
Method 1:
We can divide the isosceles triangle in two right triangles,
hypotenuse=7
side=9/2=4.5
B=A=arccossine (4.5/7)=49.994799...º≈50º
C/2=90º-50º=40º ⇒ C=2*40º=80º
Answer:
a=7; A=50º
b=7; B=50º
<span>c=9; C=80º
Method 2:
Law of cosines:
a²=b²+c²-2bcCosA ⇒CosA=(a²-b²-c²)/(-2bc)
CosA=(49-49-81) / (-126)=0.642857
A=arco cos (81/126)≈50º
B=A=50º
A+B+C=180º
50º+50º+C=180º
C=180º-100º
C=80º
Answer:
</span>a=7; A=50º
b=7; B=50º
<span>c=9; C=80º</span>
