Simple....
you have 7(5x+3)+3(12-x)....
just take them apart....
7(5x+3)
distribute the 7...
7*5x=35x
and
7*3=21
Now, the other part...
3(12-x)
Distribute the 3...
3*12=36
and
3*-x=-3x
Now, add what you have found....
35x+21+36-3x
32x+57
Thus, your answer, A.
Answer:
The answer is 108 I remember it because 9×12 is 108 and that's kinda what it is really.
Answer:
7.) 7
10.) 0
Step-by-step explanation:
When it means "evaluate the function", it's in essence asking us to see what the function spits out when we feed it a certain input. Our inputs are our x values, which spit out a y value.
Evaluating the function when x = 1:
Let's look at where the function has an x value of 1. We see it near the bottom of the table and see the y value associated with the input is 7. So when the function is fed 1 as an input, it spits out 7.
Evaluating the function when f(x) = - 2:
This one is a weird because of the new notation. Just think of it as some value of f, which we don't know (so we represent it as an x-variable) must equal -2. So let's look at our table to find out where our output is -2. We find that when f(x) = -2 the input is 0. So the input which gives -2 is 0.
Answer:
f(x) = ax^5 + bx^3 + cx^2 + 12.
Step-by-step explanation:
the degree is 5 so we will have a term like:
ax^5
it crosses the vertical axis at y = 12 so we will have a constant term equal to 12
the polynomial has 4 terms, and we already have two so we can make up two more but the exponent must be between 1 and 4
f(x) = ax^5 + bx^3 + cx^2 + 12.
it has 4 terms, f(0) = 12, then it intersects the y-axis at y = 12, and the maximum exponent is 5, then the degree of f(x) is 5
