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:
W = 16 in
Step-by-step explanation:
P = 2L + 2W
56 = 2(12) + 2W
56 = 24 + 2W
56-24 = 2W
32 = 2W
W = 32/2
W = 16 in
Best regards
I think the answer would be 99
you multiply 9 and 11 and get 99
- Simplify :- 1 + - w² + 9w.
Quadratic polynomial can be factored using the transformation 
, where 
are the solutions of the quadratic equation 
.
All equations of the form 
can be solved using the quadratic formula: 
. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
Square 9.
Multiply -4 times -1.
Add 81 to 4.
Multiply 2 times -1.
Now solve the equation 
when ± is plus. Add -9 to 
.
Divide -9+ by -2.
Now solve the equation when ± is minus. Subtract from -9.
Divide 
by -2.
Factor the original expression using 
. Substitute 
for 
and 
for 
.
<h3>NOTE :-</h3>
Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨
This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)
