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kati45 [8]
2 years ago
11

Help please on this Question

Mathematics
1 answer:
Assoli18 [71]2 years ago
3 0

Answer:

A . a 1/12

Step-by-step explanation:

you divide 1/3 by 1/4 and leave a the same and you get 0.08333 but when you turn it into a fraction it becomes 1/12 . So its a 1/12

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Your friend just purchased a new sports car for $32,000. He received $6,000 for his trade in and he used that money as a down pa
V125BC [204]
Hi there

the installment price of the loan is

619.71×48=29,746.08....answer

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5 0
2 years ago
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Help lol <br> question 1 and 2 <br> 15 points
DiKsa [7]

Answers:

  1. Choice B) y = (1/4)x
  2. The unit rate is <u>  3.89  </u> dollars per pound. It costs <u>  23.34  </u> dollars for a turkey that weighs 6 pounds.

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

Explanations:

  1. Direct proportional equations are always in the form y = kx, where k is some constant. In the case of choice B, we have k = 1/4. Visually, direct proportion equations are always straight lines and go through the origin.
  2. Divide $34.99 over 9 to get 34.99/9 = 3.88777 approximately. This rounds to 3.89, so it costs $3.89 per pound. That's the unit rate. Multiply this by 6 to get 6*3.89 = 23.34 which is the cost of the 6 pound turkey.
6 0
3 years ago
Read 2 more answers
Prove that
Pani-rosa [81]
Let's start from what we know.

(1)\qquad\sum\limits_{k=1}^n1=\underbrace{1+1+\ldots+1}_{n}=n\cdot 1=n\\\\\\&#10;(2)\qquad\sum\limits_{k=1}^nk=1+2+3+\ldots+n=\dfrac{n(n+1)}{2}\quad\text{(arithmetic  series)}\\\\\\&#10;(3)\qquad\sum\limits_{k=1}^nk\ \textgreater \ 0\quad\implies\quad\left|\sum\limits_{k=1}^nk\right|=\sum\limits_{k=1}^nk

Note that:

\sum\limits_{k=1}^n(-1)^k\cdot k^2=(-1)^1\cdot1^2+(-1)^2\cdot2^2+(-1)^3\cdot3^2+\dots+(-1)^n\cdot n^2=\\\\\\=-1^2+2^2-3^2+4^2-5^2+\dots\pm n^2

(sign of last term will be + when n is even and - when n is odd).
Sum is finite so we can split it into two sums, first S_n^+ with only positive trems (squares of even numbers) and second S_n^- with negative (squares of odd numbers). So:

\sum\limits_{k=1}^n(-1)^k\cdot k^2=S_n^+-S_n^-

And now the proof.

1) n is even.

In this case, both S_n^+ and S_n^- have \dfrac{n}{2} terms. For example if n=8 then:

S_8^+=\underbrace{2^2+4^2+6^2+8^2}_{\frac{8}{2}=4}\qquad\text{(even numbers)}\\\\\\&#10;S_8^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{8}{2}=4}\qquad\text{(odd numbers)}\\\\\\

Generally, there will be:

S_n^+=\sum\limits_{k=1}^\frac{n}{2}(2k)^2\\\\\\S_n^-=\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\\\\\\

Now, calculate our sum:

\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=&#10;\left|\sum\limits_{k=1}^\frac{n}{2}(2k)^2-\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\right|=\\\\\\=&#10;\left|\sum\limits_{k=1}^\frac{n}{2}4k^2-\sum\limits_{k=1}^\frac{n}{2}\left(4k^2-4k+1\right)\right|=\\\\\\

=\left|4\sum\limits_{k=1}^\frac{n}{2}k^2-4\sum\limits_{k=1}^\frac{n}{2}k^2+4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|=\left|4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|\stackrel{(1),(2)}{=}\\\\\\=&#10;\left|4\dfrac{\frac{n}{2}(\frac{n}{2}+1)}{2}-\dfrac{n}{2}\right|=\left|2\cdot\dfrac{n}{2}\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\left|n\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\\\\\\&#10;

=\left|\dfrac{n^2}{2}+n-\dfrac{n}{2}\right|=\left|\dfrac{n^2}{2}+\dfrac{n}{2}\right|=\left|\dfrac{n^2+n}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\\\\\\\stackrel{(2)}{=}&#10;\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk

So in this case we prove, that:

 \left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk

2) n is odd.

Here, S_n^- has more terms than S_n^+. For example if n=7 then:

S_7^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{n+1}{2}=\frac{7+1}{2}=4}\\\\\\&#10;S_7^+=\underbrace{2^2+4^4+6^2}_{\frac{n+1}{2}-1=\frac{7+1}{2}-1=3}\\\\\\

So there is \dfrac{n+1}{2} terms in S_n^-, \dfrac{n+1}{2}-1 terms in S_n^+ and:

S_n^+=\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2\\\\\\&#10;S_n^-=\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2

Now, we can calculate our sum:

\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=&#10;\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2-\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2\right|=\\\\\\=&#10;\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}\left(4k^2-4k+1\right)\right|=\\\\\\=&#10;\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}4k^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\

=\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-4\left(\dfrac{n+1}{2}\right)^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\=&#10;\left|-4\left(\dfrac{n+1}{2}\right)^2+4\sum\limits_{k=1}^{\frac{n+1}{2}}k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|\stackrel{(1),(2)}{=}\\\\\\&#10;\stackrel{(1),(2)}{=}\left|-4\dfrac{n^2+2n+1}{4}+4\dfrac{\frac{n+1}{2}\left(\frac{n+1}{2}+1\right)}{2}-\dfrac{n+1}{2}\right|=\\\\\\

=\left|-n^2-2n-1+2\cdot\dfrac{n+1}{2}\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=&#10;\left|-n^2-2n-1+(n+1)\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=&#10;\left|-n^2-2n-1+\dfrac{(n+1)^2}{2}+n+1-\dfrac{n+1}{2}\right|=\\\\\\=&#10;\left|-n^2-n+\dfrac{n^2+2n+1}{2}-\dfrac{n+1}{2}\right|=\\\\\\=&#10;\left|-n^2-n+\dfrac{n^2}{2}+n+\dfrac{1}{2}-\dfrac{n}{2}-\dfrac{1}{2}\right|=\left|-\dfrac{n^2}{2}-\dfrac{n}{2}\right|=\left|-\dfrac{n^2+n}{2}\right|=\\\\\\

=\left|-\dfrac{n(n+1)}{2}\right|=|-1|\cdot\left|\dfrac{n(n+1)}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk

We consider all possible n so we prove that:

\forall_{n\in\mathbb{N}}\quad\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk
7 0
3 years ago
Convert 0.6 shuttle orbits per hour to orbits per year
Hoochie [10]

Answer:

5256 orbits

Step-by-step explanation:

There are 8760 hours in a year, and at a rate of 0.6 orbits per hour you can find how many orbits there are per year by multiplying the rate and time like this

8760 x 0.6

This gives us an answer of 5256 which is the number of orbits per year

4 0
3 years ago
Which statement is true
weqwewe [10]
The answers are
a) the domain is all real numbers
d) the input to an exponential function is the exponent
e) the base represents the multiplicative rate of change

hope this helps :)
5 0
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