Answer: 24ft squared
Step-by-step explanation:
8x3 = 24
Answer:
See below for proof.
Step-by-step explanation:
<u>Given</u>:

<u>First derivative</u>

<u />
<u />
<u />

<u>Second derivative</u>
<u />







<u>Proof</u>



![= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%28x%2B%5Csqrt%7B1%2Bx%5E2%7D%5Cright%29%5Em%5Cleft%5Bm%5E2-%5Cdfrac%7Bmx%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%7D%2B%5Cdfrac%7Bmx%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%7D-m%5E2%5Cright%5D)
![= \left(x+\sqrt{1+x^2}\right)^m\left[0]](https://tex.z-dn.net/?f=%3D%20%5Cleft%28x%2B%5Csqrt%7B1%2Bx%5E2%7D%5Cright%29%5Em%5Cleft%5B0%5D)

Answer:
Step-by-step explanation:
While working at his neighborhood math tutoring center researching the comprehension level of the students Dion investigated that the distribution of the student test scores are normally distributed with a mean of 79.13 and a standard deviation of 6.34. What is the probability that the student scores less than 60.11
We solve using z score formula
z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
z = 60.11 - 79.13/6.34
Answer:
Annie's Orange Grove
Step-by-step explanation:
* you need to find out how much each pound of oranges cost at both orchards.
* to do so you take the total amount of money and divide it by the number of pounds you get.
1) 7.25 ÷ 20 = .36
2) 5 ÷12 = .4
* as you can see .36 is less than .4 and therefore is the better deal.
Answer:
Exact Form: 5
/4
Decimal Form: 1.25
Mixed Number Form:
1
1 -----
4
(I put it in different forms because you did not specify how you wanted your answer)
Please mark me brainliest if this helped (I need It)