Can somebody help me with this math problem?

Mathematics

2 answers:

erik [133]3 years ago

6 0

The most logical situation would be A, the hiker starts at -10 feet the increases by 10 feet.

B. is close to A. but it wouldn't explain how the hiker stops at 0 feet. It'll just explain how he decreases by 10 feet going down to -20 feet

C. would be the same way w/ B. but now the elevation is above sea level and not below

D. however might confuse you into thinking that it is the correct answer because of the hiker starting at 0 feet. D. describes the hiker at 0 feet but then describes him increasing by 10 feet.

I hope this helps

Lilit [14]3 years ago

3 0

Since the altitude OF sea level is 0, and the hiker in Death Valley is below sea level, the equation you would write would be :
-10 + 10. This is because the joker went from BELOW sea level, to the altitude OF sea level.
So, the answer should be A

You might be interested in

⚠️⚠️Please help⚠️⚠️ Pip and Ali share S785 in the ratio Pip:Ali = 4:1.<br><br> Work out Pip's share.

oksano4ka [1.4K]

<h3>Answer: $628 </h3>

==========================================================

Work Shown:

x = Ali's share

4x = Pip's share, because it is four times bigger due to the ratio 4:1

x+4x = 5x = total amount of money = 785

5x = 785

x = 785/5

x = 157 dollars is Ali's share

4x = 4*157 = 628 dollars is Pip's share

Note that 157+628 = 785 to help confirm the answer.

7 0

3 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$