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OlgaM077 [116]
3 years ago
6

Find the product of (p+3q2)pq

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Answer:

p²q + 3pq³

Step-by-step explanation:

(p + 3q²) pq ← multiply each term in the parenthesis by pq

= p²q + 3pq³

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Giving 100 points.
Nitella [24]

Answer:

1.   <u>Cost per customer</u>:  10 + x

     <u>Average number of customers</u>:  16 - 2x

\textsf{2.} \quad  -2x^2-4x+160\geq 130

3.    $10, $11, $12 and $13

Step-by-step explanation:

<u>Given information</u>:

  • $10 = cost of buffet per customer
  • 16 customers choose the buffet per hour
  • Every $1 increase in the cost of the buffet = loss of 2 customers per hour
  • $130 = minimum revenue needed per hour

Let x = the number of $1 increases in the cost of the buffet

<u>Part 1</u>

<u></u>

<u>Cost per customer</u>:  10 + x

<u>Average number of customers</u>:  16 - 2x

<u>Part 2</u>

The cost per customer multiplied by the number of customers needs to be <u>at least</u> $130.  Therefore, we can use the expressions found in part 1 to write the <u>inequality</u>:

(10 + x)(16 - 2x)\geq  130

\implies 160-20x+16x-2x^2\geq 130

\implies -2x^2-4x+160\geq 130

<u>Part 3</u>

To determine the possible buffet prices that Noah could charge and still maintain the restaurant owner's revenue requirements, solve the inequality:

\implies -2x^2-4x+160\geq 130

\implies -2x^2-4x+30\geq 0

\implies -2(x^2+2x-15)\geq 0

\implies x^2+2x-15\leq  0

\implies (x-3)(x+5)\leq  0

Find the roots by equating to zero:

\implies (x-3)(x+5)=0

x-3=0 \implies x=3

x+5=0 \implies x=-5

Therefore, the roots are x = 3 and x = -5.

<u>Test the roots</u> by choosing a value between the roots and substituting it into the original inequality:

\textsf{At }x=2: \quad -2(2)^2-4(2)+160=144

As 144 ≥ 130, the <u>solution</u> to the inequality is <u>between the roots</u>:  

-5 ≤ x ≤ 3

To find the range of possible buffet prices Noah could charge and still maintain a minimum revenue of $130, substitute x = 0 and x = 3 into the expression for "cost per customer.  

[Please note that we cannot use the negative values of the possible values of x since the question only tells us information about the change in average customers per hour considering an <em>increase </em>in cost.  It does not confirm that if the cost is reduced (less than $10) the number of customers <em>increases </em>per hour.]

<u>Cost per customer</u>:  

x =0 \implies 10 + 0=\$10

x=3 \implies 10+3=\$13

Therefore, the possible buffet prices Noah could charge are:

$10, $11, $12 and $13.

8 0
1 year ago
Hey PLEASE HELP DUE TODAY
nadya68 [22]
What the maning of that ?
4 0
3 years ago
A certain car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year
icang [17]

Answer:

b(b/a)^2

Step-by-step explanation:

Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then

b = a - (p% × a) = a(1-p%)

b/a = 1 - p%

p% = 1 - b/a = (a-b)/a

Let the worth of the car on December 31, 2012 be c

then

c = b - (b × p%) = b(1-p%)

Let the worth of the car on December 31, 2013 be d

then

d = c - (c × p%)

d = c(1-p%)

d = b(1-p%)(1-p%)

d = b(1-p%)^2

d = b(1- (a-b)/a)^2

d = b((a-a+b)/a)^2

d = b(b/a)^2 = b^3/a^2

The car's worth on December 31, 2013 =  b(b/a)^2 = b^3/a^2

4 0
3 years ago
Combine the following fractions by addition or subtraction as directed.
Marina86 [1]
Answer:\frac{6}{y^2-xy}-\frac{6}{x^2-xy}=\frac{6(x+y)}{xy(y-x)}

Explanation:

Combining the fractions, we have:

\begin{gathered} \frac{6(x^2-xy)-6(y^2-xy)}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6xy-6y^2+6xy}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6y^2}{x(x-y).y(y-x)} \\  \\ =\frac{6(x-y)(x+y)}{-xy(x-y)^2} \\  \\ =\frac{-6(x+y)}{xy(x-y)} \\  \\ =\frac{6(x+y)}{xy(y-x)} \end{gathered}

8 0
1 year ago
Solve.<br>6x2 + 7x – 20 = 0​
Oliga [24]

Answer:

(3x + 4)(2x - 5)

6x^{2}+7x-20=0<em>    =    (3x + 4)(2x - 5)</em>

Hope this helped! Have a great day!

3 0
2 years ago
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