The ounces of chocolate chips used by Mrs Jacob is 70 ounce
<em><u>Solution:</u></em>
Given that Jacob is making several batches of cookies and is using 84 total ounces of chips
Let "c" be the ounces of chocolate chips
Let "p" be the ounces of peanut butter chips
To find: ounces of chocolate chips used by Mrs Jacob
Given that There are 5 times as many ounces of chocolate chips as peanut butter chips
<em><u>Thus we can frame a equation as:</u></em>
ounces of chocolate chips = 5 x ounces of peanut butter chips
c = 5p -------- eqn 1
Jacob used 84 total ounces of chip. Therefore,
ounces of chocolate chips + ounces of peanut butter chips = 84
c + p = 84 ---- eqn 2
Substitute eqn 1 in eqn 2
5p + p = 84
6p = 84
<h3>p = 14</h3>
Substitute p = 14 in eqn 1
c = 5(14) = 70
<h3>c = 70</h3>
Thus the ounces of chocolate chips used by Mrs Jacob is 70 ounce
Answer:
no because she would get the same total if she added it before or after
Step-by-step explanation:
Answer:
y = (11x + 13)e^(-4x-4)
Step-by-step explanation:
Given y'' + 8y' + 16 = 0
The auxiliary equation to the differential equation is:
m² + 8m + 16 = 0
Factorizing this, we have
(m + 4)² = 0
m = -4 twice
The complimentary solution is
y_c = (C1 + C2x)e^(-4x)
Using the initial conditions
y(-1) = 2
2 = (C1 -C2) e^4
C1 - C2 = 2e^(-4).................................(1)
y'(-1) = 3
y'_c = -4(C1 + C2x)e^(-4x) + C2e^(-4x)
3 = -4(C1 - C2)e^4 + C2e^4
-4C1 + 5C2 = 3e^(-4)..............................(2)
Solving (1) and (2) simultaneously, we have
From (1)
C1 = 2e^(-4) + C2
Using this in (2)
-4[2e^(-4) + C2] + 5C2 = 3e^(-4)
C2 = 11e^(-4)
C1 = 2e^(-4) + 11e^(-4)
= 13e^(-4)
The general solution is now
y = [13e^(-4) + 11xe^(-4)]e^(-4x)
= (11x + 13)e^(-4x-4)
Answer:
x = 7
Step-by-step explanation:
I hope this helps!
<em>Answer:</em>
24.64
<em>Method:</em>
You get 10% of the shirt cost, and it on to the 22.40
