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yKpoI14uk [10]
3 years ago
8

A company earns a weekly profit of P dollars by selling x items, according to the equation P(x)=-5x^2+40x-300. How many items do

es the company have to sell each week to maximize the profit?
Mathematics
1 answer:
STALIN [3.7K]3 years ago
3 0
Since the equation is:

<span>P(x)=-5x^2+40x-300

Solving the equation by using the calculator, the values for x are:

x = 4.72

Therefore, the company must sell 8 items per week to maximize the profit.

</span>I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Step-by-step explanation:

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Urgent question in Mathematics. Solve this to get 15 points
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Answer:

1.

Step-by-step explanation:

Note that 2^4 - 2^3 = 16 - 8 = 8 = 2^3

In a similar way 2^6 - 2^5 = 64 - 32 = 32 = 2^5.

So 2^99 - 2^98 = 2^98 ,  2^98 - 2^97 = 2^97 ,  2^97 - 2^96 = 2^96 and so on.

Therefore when we come to the last 2 terms we have 2^1 - 2^0 = 2 - 1

= 1 , so the answer is 1.

3 0
3 years ago
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Prove for any positive integer n, n^3 +11n is a multiple of 6
suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

----------------------------------

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

-----------------------------------

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

------------------------------------

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

-------------------------------------

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

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saw5 [17]
The x is going to be a negative number to get the 3. so start off minus 5 from plus 5 and minus five from the other side too. the positive 5 and the negative cancel out and then take it fromt here with three and the negative 5. 
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Find the hypotenuse of a right triangle that has one side of 7 cm and another<br> one of 3 cm.
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The hypotenuse is the square root of 58.
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