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Bad White [126]
2 years ago
5

PLEASE HELP ME WITH THIS PROBLEM:

Mathematics
1 answer:
Andru [333]2 years ago
3 0

Answer:

(4,3)

Step-by-step explanation:

The solution of two lines is where they intersect

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Widget Wonders produces widgets. They
navik [9.2K]
Find the total cost of producing 5 widgets. Widget Wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $15.50 to produce 3 widgets, $23.50 to produce 7 widgets, and $56 to produce 12 widgets.
OK...so we have
a(7)^2 + b(7) + c = 23.50 → 49a + 7b + c = 23.50 (1)
a(3)^2 + b(3) + c = 15.50 → 9a + 3b + c = 15.50 subtracting the second equation from the first, we have
40a + 4b = 8 → 10a + b = 2 (2)
Also
a(12)^2 + b(12) + c = 56 → 144a + 12b + c = 56 and subtracting (1) from this gives us
95a + 5b = 32.50
And using(2) we have
95a + 5b = 32.50 (3)
10a + b = 2.00 multiplying the second equation by -5 and adding this to (3) ,we have
45a = 22.50 divide both sides by 45 and a = 1/2 and using (2) to find b, we have
10(1/2) + b = 2
5 + b = 2 b = -3
And we can use 9a + 3b + c = 15.50 to find "c"
9(1/2) + 3(-3) + c = 15.50
9/2 - 9 + c = 15.50
-4.5 + c = 15.50
c = 20
So our function is
c(x) = (1/2)x^2 - (3)x + 20
And the cost to produce 5 widgets is = $17.50
7 0
3 years ago
"Tim's lawn is 20 meters by 30 meters. If Tim can mow 8 square meters of grass per minute, how long will it take him to mow the
frosja888 [35]
One hour and a half
5 0
3 years ago
5,000 tens equals how many thousands
Brilliant_brown [7]

5,000 tens = (5,000 x 10)

To find out how many thousands are in it,
just divide:

(5,000 x 10) / (1,000)  =  (5 x 10)  =  50 of them .

4 0
3 years ago
IS THIS CORRECT??????????
Alika [10]
The correct answers are the 3rd and 4th ones.
7 0
3 years ago
Read 2 more answers
Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con
monitta

Let f(x)=e^{-1/x}. Then f'(x)=\frac1{x^2}e^{-1/x}>0 for all x\ge2, so f is strictly increasing. As x\to\infty, e^{-1/x}\to e^0=1, so f is bounded above by 1. This is to say,

e^{-1/x}

and the integral of \frac1{x^2} converges over the same domain, so this integral must also converge by comparison.

We have, by setting y=-\frac1x,

\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}

8 0
3 years ago
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