Answer: f(2) = -4
Step-by-step explanation: Here we're given the function f(x) = -3x + 2 and we are asked to find f(2).
In other words, if we put an <em>x</em> into our function, we get a -3x + 2 out so what happens when we put a 2 into the function?
Well if we put a 2 into the function,
that's f(2) and we get -3(2) + 2 out.
Now all we have to do is simplify on the right side.
-3(2) is -6 so we have -6 + 2 which is -4.
So f(2) is -4.
Answer:
could you possibly zoom in
Step-by-step explanation:
I also worked it out but I got -5/3
2(6p-5) ≥ 3(p-8) -1
12p-10 ≥ 3p-24-1
12p-10 ≥ 3p-25
12p-3p ≥ -25+10
9p ≥ -15
-15/9
P ≥ -5/3
Answer:
473 yd^3
Step-by-step explanation:
The surface area of the square base is (11 yd)^2, or 121 yd^2.
The surface area of one of the slanting sides of the pyramid is
A = (1/2)(b)(h), which comes to A = (1/2)(11 yd)(16 yd) = 88 yd^2.
Thus, four such sides have a combined surface area of 352 yd^2.
Combining the surface areas of the slanting sides with the base area yields 473 yd^3
Answer:
What is the graph of h(x)=f(x)+g(x) with an example?
So many possible combinations of types of equations for f(x) and g(x).
If they are both linear. f(x) = 3x + 2. g(x) = 2x - 5. h(x) = f(x) + g(x) = 5x - 3. This is also linear.
f(x) has slope = 3 and y-intercept = 2. g(x) has slope = 2 and y intercept = -5. h(x) has slope = 5 and y-intercept = -3.
The graph of the sum of two linear equations is a straight line with slope equal to the sum of the slopes of the two linear equations and a y-intercept equal to the sum of the y-intercepts of the two linear equations.
If one is linear and the other is quadratic. f(x) = 2x + 3. g(x) = x^2 + 6x - 4. h(x) = f(x) + g(x) = x^2 + 8x - 1. This is quadratic.
f(x) has slope = 3 and y-intercept = 3. g(x) has an axis of symmetry of x = -3, vertex at (-3, -13), y-intercept = -4, x-intercepts = -3 + 13^½ and -3 - 13^½ . h(x) has an axis of symmetry of x = -4, vertex at (-4, -17), y-intercept = -1, x-intercepts = -4 + 17^½ and -4 - 17^½ .
The graph of the sum of a linear equation [y = mx + b] and a quadratic equation [y = Ax^2 + Bx + C] has an axis of symmetry of x = - (B + m) / 2A, vertex at ( - (B + m) / 2A, - (B + m)^2 / 4A + (b + C)), y-intercept = b + C, x-intercepts = (- (B + m) + ( (B + m)^2 - 4A (b + C))^½ ) / 2A and (- (B + m) - ( (B + m)^2 - 4A (b + C))^½ ) / 2A .