G(x)=-2x-4
f(x)=15+0.75x
the common solution Q is their intersection-> calculate x by setting them equal:
-2x-4=15+0.75x
-2x-(3/4)x=19
-(11/4)x=19
x=-4*19/11
x=-76/11
calculate y:
f(-76/11)=15+(3/4)*(-76/11)
=15+(-3*19/11)
=15+(-57/11)
=15+-5-(2/11)
=10-(2/11)
=108/11
so the solution is a) (-76/11, 108/11)
Answer:
48390,"37,520","65,810",2170 or 2180 (both will work)
180 I’m pretty sure look 1x10x100(cents)x18(dollars)/100 if it’s not that if it’s a not a pair of questions then it’s either 180 or 1,800
f(3) = 69
Step-by-step explanation:
Given f(x) =5x5 + 10x4 + 3x3 + 8x2 – 6x – 3; c = 3
f(x) = 25+40+9+16 - 6x - 3
f(x) = - 6x +90-3
f(x) = - 6f(3) = 69x + 87
Taken c = 3
f(c) = - 6c +90-3
f(c) = - 6c + 87
substitute c=3
f(3) = - 6 (3) +90-3
= -18 +87
= 69
Hence .,f(3) = 69
Answer:
<h2> 105 tickets</h2>
Step-by-step explanation:
To solve this problem we need to model an equation to represent the situation first.
the goal is to archive $7500 in the even, bearing in mind that there is a cost of $375 fee for rent, we need to put this amount into consideration
let the number of tickets be x
so
75x-375>=7500--------1
Equation 1 above is a good model for the equation
we can now solve for x to determine the number of tickets to be sold to archive the aim
75x-375>=7500--------1
75x>=7500+375
75x>=7875
divide both sides by 75 we have
x>=7875/75
x>=105 tickets
so they must sell a total of 105 tickets and above to meet the target of $7500 with the rent inclusive
