Amount of money paid by Jeremy as rent and maintenance of shop per month = $1500
Cost of raw materials and manufacturing per month = $6000
Total cost that Jeremy has to spend per month = (1500 + 6000) dollars
= 7500 dollars
Number of individual chocolates sold = 2400
Number of chocolates sold in boxes = 50 boxes
= (12 * 50) chocolates
= 600 chocolates
Then
Total number of chocolates sold by Jeremy = 2400 + 600
= 3000
Now
Price of each chocolate = 7500/3000 dollars
= 75/30 dollars
= 5/2 dollars
= 2.5 dollars
Price of 600 chocolates = 600 * (5/2) dollars
= 300 * 5 dollars
= 1500 dollars
Price of 12 chocolates = (1500/600) * 12
= 15 * 2
= 30 dollars
Then
We can say that each box of chocolate should be sold at $30. All the loose chocolates should be sold at $2.5 each.
Answer:
F
Step-by-step explanation:
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The answer is B.
The formula for arc length is s = r(theta) Theta must be in radians, so convert 40 degrees to radians, which is 2pi/9. Multiply 2pi/9 by the radius, 9, and then you'll get the answer. Hopes this helps!
Answer:
a = 0.25
Step-by-step explanation:
Our strategy to solve this problem will be to use the information given in the table to obtain first the value of c in the quadratic equation which has the form ax^2 + bx + c and then form a system of 2 linear equations and solve for the coefficients a and b as follow:
from x=0 and y=-3 a*0 + b* 0 + c = - 3
c = -3
from x=1 and y= -3.75
a*(1) + *(1) + (-3) = -3.75
a + b = -3.75 + 3 = -0.75
from x= 2 and y = -4
a* (2)^2 + b*(2) + (-3) = -4
4a + 2b = -4 +1
4a + 2b = -1
Now we can solve the system f equations by elimination:
a + b = -0.75
4a + 2b = -1
multiply first equation by -2 and add to the second and get
-2a - 2b = +1.50
4a + 2b = -1
2a = 0.50 and substituting into any of the equations get b = -1
so the quadratic equation has a= 0.25 b= -1 c= -3
we can even plug the any of the other values for x given in the table and check the answer.
System 1: The solution is (x, y) = (-4, 5)
System 2: The solution is 
<em><u>Solution:</u></em>
<em><u>Given system of equations are:</u></em>
2x + 3y = 7 ------ eqn 1
-3x - 5y = -13 --------- eqn 2
We can solve by elimination method
Multiply eqn 1 by 3
6x + 9y = 21 ------ eqn 3
Multiply eqn 2 by 2
-6x - 10y = -26 ------- eqn 4
Add eqn 3 and eqn 4
6x + 9y -6x - 10y = 21 - 26
-y = -5
y = 5
Substitute y = 5 in eqn 1
2x + 3(5) = 7
2x + 15 = 7
2x = -8
x = -4
Thus the solution is (x, y) = (-4, 5)
<h3><em><u>
Second system of equation is:</u></em></h3>
8 - y = 3x ------ eqn 1
2y + 3x = 5 ----- eqn 2
We can solve by susbtitution method
From given,
y = 8 - 3x ----- eqn 3
Substitute eqn 3 in eqn 2
2(8 - 3x) + 3x = 5
16 - 6x + 3x = 5
3x = 16 - 5
3x = 11
Substitute the above value of x in eqn 3
y = 8 - 3x
Thus the solution is
