Answer:
126
Step-by-step explanation:
Although I don't know what you mean by partial products here is a step by step.
42*3
120+6
126
Since NO=NP (as seen by the 2 lines next to each other), we get
17=5x-6. Adding 6 to both sides, we get 23=5x and by dividing 5 on both sides we get x=23/5
E
Given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 ), then
The x-coordinate of the vertex is
= - 
y = 10x² + 5x - 7 is in standard form
with a = 10, b = 5 and c = - 7, hence
= - 
= - 
Substitute this value into the equation for y-coordinate
y = 10 (- )² + 5 (- ) - 7
= 
- 
- 7
= - 
- 
= - 
vertex = (- 0.25, - 7.625 ) → E
11
first you have to read the graph to see what f(x) is. You know that when x = 0 then y= - 1 so what you have so far is y = ax^2 - 1. You have to guess a bit to see what a is.
Just guessing (and that's all you can do unless you have more information to the question) it looks like when y = 0 then x = about 1.4 or -1.4. That's a good number to have.
y = 1/2 x^2 - 1 is probably the original equation that produced the graph.
So let's check it out.
0 = 1/2 x^2 - 1 Add 1 to both sides.
1 = 1/2 x^2 Multiply by 2
2 = x^2 Take the square root of both sides.
sqrt(2) = sqrt(x^2)
x = +/- sqrt(2)
x = +/- 1.4142 about. So our graph is a pretty good approximation of y = (1/2)x^2 - 1
Now what you want is f(x).
y and f(x) are very close to being equal so we have f(x) = 1/2 x^2 - 1
Now you can solve for f(1)
f(1) = 1/2 (1)^2 - 1
f(1) = 1/2 (1) - 1
f(1) = 1/2 - 1
f(1) = - 1/2 <<<<< Answer.
Answer:
r(6)/s(6) = 3(6) -1
--------------
2(6) +1
Step-by-step explanation:
r(x) = 3x-1
Let x=6
r(6) = 3(6) -1
s(x) = 2x+1
Let x=6
s(6) = 2(6) +1
r(6)/s(6) = 3(6) -1
--------------
2(6) +1
