A.
F(10)= (10)^2-3(10)-5
F(10)=100-30-5
F(10)=65
B.
F(-3)=(-3)^2-3(-3)-5
F(-3)=9+9-5
F(-3)=13
C.
G(2)= -6(2)+1
G(2)= -12+1
G(2)=-11
D.
G(10)=-6(10)+1
G(10)= -60+1
G(10)= -59
Answer:
Step-by-step explanation:
Answer:
1. -7.5
2. $1
3. 40
Step-by-step explanation:
For number 1, it can be solved by using the PEMDAS method, or see explanation below:
6x - 4x - 36 = 6 - 2x
2x - 36= -2x + 6
4x + 36 = 6
4x = -30
x = -15/2 or -7.5
For number 2, substitute 3 into both equations:
f(x) =1.50(3) + 2.00
and
f(x) = 2.00(3) + 1.50
This would get $6.50 and $7.50, which, if subtracted, gets $1.
For number 3, do something similar to the previous problem. Substitute 3 for x. It would be 5(2^3), or 40.
Hope this helps!
Answer:
<h2>360 cakes</h2>
Step-by-step explanation:
<h2>soln:</h2><h2>onecake =240\8=30</h2><h2>then 12 cakes =30×12=360</h2>
Answer:
<h2>absolute maximum = 16</h2><h2>absolute minimum = 1</h2>
Step-by-step explanation:
To get the absolute maximum and minimum values of the function f(x) = 16 + 2x − x² n the given interval [0,5], we need to get the values of f(x) at the end points. The end points are 0 and 5.
at x = 0;
f(0) = 16 + 2(0) − 0²
f(0) = 16
at the other end point i.e at x = 5;
f(5) = 16 + 2(5) − 5²
f(5) = 16 + 10-25
f(5)= 26-25
f(5) = 1
The absolute minimum value is 1 and occurs at x = 5
The absolute maximum value is 16 and occurs at x = 0
