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gogolik [260]
3 years ago
11

How do you find the area with only this information?

Mathematics
1 answer:
just olya [345]3 years ago
8 0
I would go with J as your answer
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64 ÷ 13632 can u help me
denis-greek [22]

Answer:

it is 0.00469483568



4 0
3 years ago
Read 2 more answers
If
Nitella [24]

Since profit can't be negative, the production level that'll maximize profit is approximately equal to 220.

<h3>How to find the production level that'll maximize profit?</h3>

The cost function, C(x) is given by 12000 + 400x − 2.6x² + 0.004x³ while the demand function, P(x) is given by 1600 − 8x.

Next, we would differentiate the cost function, C(x) to derive the marginal cost:

C(x) = 12000 + 400x − 2.6x² + 0.004x³

C'(x) = 400 − 5.2x + 0.012x².

Also, revenue, R(x) = x × P(x)

Revenue, R(x) = x(1600 − 8x)

Revenue, R(x) = 1600x − 8x²

Next, we would differentiate the revenue function to derive the marginal revenue:

R'(x) = 1600 - 8x

At maximum profit, the marginal revenue is equal to the marginal cost:

1600 - 8x = 400 − 5.2x + 0.012x

1600 - 8x - 400 + 5.2x - 0.012x² = 0

1200 - 2.8x - 0.012x² = 0

0.012x² + 2.8x - 1200 = 0

Solving by using the quadratic equation, we have:

x = 220.40 or x = -453.73.

Since profit can't be negative, the production level that'll maximize profit is approximately equal to 220.

Read more on maximized profit here: brainly.com/question/13800671

#SPJ1

6 0
3 years ago
If a piano weighs 289 kilograms more than the piano bench that weighs 297kg. How much does the piano bench weighs?
juin [17]
586 kg is the answer

8 0
4 years ago
Read 2 more answers
WELP ITS MATH SEE BELOW
never [62]

Answer:

26/52

Step-by-step explanation:

I believe this is correct

6 0
4 years ago
Find the length of the curve given by ~r(t) = 1 2 cos(t 2 )~i + 1 2 sin(t 2 ) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplif
xxMikexx [17]

Answer:

The length of the curve is

L ≈ 0.59501

Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

8 0
3 years ago
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