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Luda [366]
3 years ago
10

Devin grew 12 flowers with 3 seed packets. How many seed packets does Devin need to have a total of 20 flowers in his garden? As

sume the relationship is directly proportional.​
Mathematics
1 answer:
Whitepunk [10]3 years ago
7 0
If he grew 12 flowers with 3 seed packets then there should be 4 seeds in one packet > 3 x 4 = 12
Devin needs 5 seed packets to have a total of 20 flowers in his garden
> 4 x 5 = 20
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A company’s profit in dollars is modeled by the equation p(x) = –0.5x2 + 100x, where x represents the number of units sold.
serg [7]

Answers:

a. The company needs to sell 100 units to reach the maxmum profit.

b. The company's maximum profit is $5,000.

Solution:

a. How many unit does the company need to sell to reach the maximum profit?

p(x)=-0.5x^2+100x

This is a quadratic equation, and its graph is a parabola vertical (because the veriable "x" is square). Comparing with the general form:

p(x)=ax^2+bx+c; a=-0.5, b=100, c=0

a=-0.5<0 (negative), then the parabola opens downward, and it has a maximimun value (maximum profit) at its vertex.

We can find the abscissa of the vertex (units that the company needs to sell to reach the maximum profit using the following formula:

x=-b/(2a)

Replacing b by 100 and a by -0.5 in the formula above:

x=-100/[2(-0.5)]

x=-100/(-1)

x=100

The company needs to sell 100 units to reach the maximum profit.


b. What's the company's maximum profit?  

To determine the maximum profit we substitute the value of "x" obrained in part "a" in the quadratic equation:

x=100→p(100)=-0.5(100)^2+100(100)

p(100)=-0.5(10,000)+10,000

p(100)=-5,000+10,000

p(100)=5,000

The company's maximum profit is $5,000.

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Jim's father bought a new car 3 years ago for
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Solve the system of linear equations below.
Olegator [25]

Answer:

B. x = 2, y = 6

Step-by-step explanation:

2x + 5y = 34...(1) \\  \\ x + 2y = 14 \\  \therefore \: x = 14 - 2y...(2) \\  \\ from \: equations \: (1) \: and \: (2) \\  \\ 2(14 - 2y) + 5y = 34 \\ \therefore \: 28 - 4y + 5y = 34 \\  \therefore \: 28 + y = 34 \\ \therefore \:  y = 34 - 28 \\  \huge \red{ \boxed{\therefore \:  y = 6}} \\ substituting \: y = 6 \: in \: equation \: (2) \\ x = 14 - 2 \times 6 \\ \therefore \:  x =14 - 12 \\  \huge \purple{ \boxed{\therefore \:  x =2}} \\  \\ \huge \orange{ \boxed{ \therefore \:  x  = 2, \:  \: y = 6}} \\

5 0
3 years ago
Read 2 more answers
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