Answer:
Wonka bars=3 and Everlasting Gobstoppers=24
Step-by-step explanation:
let the wonka bars be X
and everlasting gobstoppers be Y
the objective is to
maximize 1.3x+3.2y=P
subject to constraints
natural sugar
4x+2y=60------1
sucrose
x+3y=75---------2
x>0, y>0
solving 1 and 2 simultaneously we have
4x+2y=60----1
x+3y=75------2
multiply equation 2 by 4 and equation 1 by 1 to eliminate x we have
4x+2y=60
4x+12y=300
-0-10y=-240
10y=240
y=240/10
y=24
put y=24 in equation 2 we have'
x+3y=75
x+3(24)=75
x+72=75
x=75-72
x=3
put x=3 and y=24 in the objective function we have
maximize 1.3x+3.2y=P
1.3(3)+3.2(24)=P
3.9+76.8=P
80.7=P
P=$80.9
1.
<span>(x-4)^2 -28 = 8
</span>(x-4)^2 = 8 + 28 = 36
<span>(x-4) = +/-6
x = 6 + 4 or -6 + 4
x = 10 or -2
2.
Mistake in </span><span>Step 5: x – 3 = 16
</span><span>Step 4: (x – 3)² = 16
</span><span>Step 5: (x – 3) = +/-4
Step 6: x = 4 + 3 or -4 + 3
Step 7: x = 7 or -1
3.
</span><span>x^2 - 4x + 3 = 0
</span><span>quadratic eqn x = [-b +/- sqrt(b^2-4ac)]/2a
where in this case
a = 1, b = -4, c = 3
x = </span> [-(-4) +/- sqrt((-4)^2-4*1*3)]/2*1
= [4 +/- sqrt(16 - 12)]/2
<span>= [4 +/- sqrt(4)]/2
= [4 +/- 2]/2
= 6/2 or 2/2
= 3 or 1
</span>
1.1x+1.2x-5.4=-10
11/10x+12/10x-54/10=10
11x/2.5+2^2-1*3/5x-3^3/5=-10
(11x)+2(2*3x)+2(-3^3)/2*5=-10
11x+2(6x)+2(-27)/2*5=-10
11x+2*6x-2*27/2*5=-10
11x+12x-54/2*5=-10
23x-54/2*5=-10
23x-54=-100
23x=-46
23x/23=-46/23
x=-2*23/23
x=-2