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Natalija [7]
3 years ago
7

The length of a rectangle is 6m longer than its width. If the perimeter of the rectangle is 52m, find its area

Mathematics
1 answer:
sveticcg [70]3 years ago
4 0
If length is x, then width is 6+x 
<span>Perimeter = 2*(x + 6+x) = 52 </span>
<span>2*6 + 2*2x = 52 </span>
<span>4x = 52 - 12 = 40 </span>
<span>x = 40/4 = 10 </span>

<span>Area = length*width = 10*(6+10) </span>
<span>Area = 10*16 = 160 meter square</span>
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Answer:

1/18 hr per stretch

Step-by-step explanation:

total # of hrs / # of stretches

1/3 hr / 6 stretches = 1/18 hr per stretch

7 0
3 years ago
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Solve x2−4x−12=0 by completing the square. A x=−2 B x=−6 or x=2 C x=−2 or x=6 D x=−6
MrRissso [65]

Answer:

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Step-by-step explanation:

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2 years ago
Could you help me to solve the problem below the cost for producing x items is 50x+300 and the revenue for selling x items is 90
s344n2d4d5 [400]

Answer:

Profit function: P(x) = -0.5x^2 + 40x - 300

profit of $50: x = 10 and x = 70

NOT possible to make a profit of $2,500, because maximum profit is $500

Step-by-step explanation:

(Assuming the correct revenue function is 90x−0.5x^2)

The cost function is given by:

C(x) = 50x + 300

And the revenue function is given by:

R(x) = 90x - 0.5x^2

The profit function is given by the revenue minus the cost, so we have:

P(x) = R(x) - C(x)

P(x) = 90x - 0.5x^2 - 50x - 300

P(x) = -0.5x^2 + 40x - 300

To find the points where the profit is $50, we use P(x) = 50 and then find the values of x:

50 = -0.5x^2 + 40x - 300

-0.5x^2 + 40x - 350 = 0

x^2 - 80x + 700 = 0

Using Bhaskara's formula, we have:

\Delta = b^2 - 4ac = (-80)^2 - 4*700 = 3600

x_1 = (-b + \sqrt{\Delta})/2a = (80 + 60)/2 = 70

x_2 = (-b - \sqrt{\Delta})/2a = (80 - 60)/2 = 10

So the values of x that give a profit of $50 are x = 10 and x = 70

To find if it's possible to make a profit of $2,500, we need to find the maximum profit, that is, the maximum of the function P(x).

The maximum value of P(x) is in the vertex. The x-coordinate of the vertex is given by:

x_v = -b/2a = 80/2 = 40

Using this value of x, we can find the maximum profit:

P(40) = -0.5(40)^2 + 40*40 - 300 = $500

The maximum profit is $500, so it is NOT possible to make a profit of $2,500.

3 0
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1 I think. Tell me if that is correct
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