Answer:
The company needs to sell either 30 or 40 items.
Step-by-step explanation:
We are given that the cost for selling <em>x</em> items given by the function:
And the revenue for selling <em>x</em> items is given by:
The profit function is the cost function subtracted from the revenue function:
Substitute and simplify:
To find how many items must be sold in order to obtain a weekly profit of $300, we can let <em>P</em> equal 300 and solve for <em>x</em>. So:
Solve for <em>x</em>. Subtract 300 from both sides:
We can divide both sides by -0.4:
Factor:
Zero Product Property:
Solve for each case:
So, in order to obtain a weekly profit of $300, the company need to sell either 30 <em>or</em> 40 items.
Step-by-step explanation:
Given the average of 8 numbers = 56.
Then the Total sum of 8 numbers = 56 * 8
= 448.
Sum of 1st three numbers = 49 + 57 + 72
= 178.
So, the total = 448 - 178
= 270.
Average of the other 5 numbers = 270/5
= 54.
Verification:
178 + 270 = 448/8
= 56.
Answer:2+x=15-7/8x
We move all terms to the left:
2+x-(15-7/8x)=0
Domain of the equation: 8x)!=0
x!=0/1
x!=0
x∈R
We add all the numbers together, and all the variables
x-(-7/8x+15)+2=0
We get rid of parentheses
x+7/8x-15+2=0
We multiply all the terms by the denominator
x*8x-15*8x+2*8x+7=0
Wy multiply elements
8x^2-120x+16x+7=0
We add all the numbers together, and all the variables
8x^2-104x+7=0
a = 8; b = -104; c = +7;
Δ = b2-4ac
Δ = -1042-4·8·7
Δ = 10592
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
x1=−b−Δ√2ax2=−b+Δ√2a
The end solution:
Δ−−√=10592−−−−−√=16∗662−−−−−−−√=16−−√∗662−−−√=4662−−−√
x1=−b−Δ√2a=−(−104)−4662√2∗8=104−4662√16
x2=−b+Δ√2a=−(−104)+4662√2∗8=104+4662√16
Step-by-step explanation:
Answer:
im pretty sure the answer is TRAPEZOID.
Step-by-step explanation:
