First simpliify
(3r^2+4)(2r-1)=
(3r^2+4)(2r)+(3r^2+4)(-1)=
(6r^3+8r)+(-3r^2-4)=
6r^3+8r-3r^2-4=
put back
6r^3-3r^2+8r-4-8r^3+6r^2=
-2r^3+3r^2+8r-4
Answer:
D
Step-by-step explanation:
A is wrong because on the point -4, the function is equal to -12 and a states the local minimum on [-4, -2] is equal to zero. Since -12 < 0, A is wrong.
B is wrong because the function never even reaches 25 on the interval [-2, -2]
C is wrong because we can see that 3.9 is zero and the function is decreasing there. fro this we can determine the functions value at x = 4 is less than 0, meaning the minimum value is also less than 0.
Lastly, we can see that the lowest point fro 4 to 7 is at -7, meaning that D is correct.
Answer: x = 14 and y = 10
Step-by-step explanation: What we have in the question is a pair of simultaneous equations as follows(
x = y + 4 ———(1)
2x - 3y = -2 ——(2)
Looking at both equations, one of the variables has a coefficient of 1, and as a result of that, we shall use the substitution method.
From equation (1), x = y + 4
Substitute for the value of x in equation (2)
2x - 3y = -2
2(y + 4) - 3y = -2
2y + 8 -3y = -2
-y + 8 = -2
Subtract 8 from both sides of the equation
-y +8 -8 = -2 -8
-y = -10
Divide both sides of the equation by -1
y = 10
Having calculated y, substitute for the value of y into equation (1)
x = y + 4
x = 10 + 4
x = 14
Therefore, x = 14 and y = 10
<span>3x(1.05)
answer
</span><span>B: 1.05x + 1.05x + 1.05x</span>