I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).
Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system
- y + 3z = -2
y + z = -2
-----------------
4z = -4, so z = -1.
Next, multiply the 3rd equation by 2: You'll get -2x + 2y + 2z = -2.
Add this result to the first equation. The 2x terms will cancel, leaving you with the system
2y + 2z = -2
y + z = 4
This would be a good time to subst. -1 for z. We then get:
-2y - 2 = -2. Then y must be 0. y = 0.
Now subst. -1 for z and 0 for y in any of the original equations.
For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.
Then a tentative solution is (-3, -1, 0).
It's very important that you ensure that this satisfies all 3 of the originale quations.
k = 5
the equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k ) are the coordinates of the vertex and a is a multiplier
To obtain this form use the method of completing the square
Since the coefficient of the x² term is 1 then
add/ subtract (half the coefficient of the x-term )² to x² - 6x
f(x) = x² + 2(- 3)x + 9 - 9 + 14 = (x - 3)² + 5 → k = 5
Answer:
Wheres the question
Step-by-step explanation:
Answer:
3. C=4a
Step-by-step explanation:
600/150 = 4 (one square foot = $4)
Answer:
6,5 5,6 3,10 10,3 1,30 30,1 2,15 15,2 = 8 possibilities
Step-by-step explanation:
