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ludmilkaskok [199]
4 years ago
10

Solve for w in terms of v, x, and y. v= -

="\frac{1}{X}" alt="\frac{1}{X}" align="absmiddle" class="latex-formula"> YW

W=
Mathematics
1 answer:
AlekseyPX4 years ago
5 0

Answer:

W = - \frac{vx}{Y}

Step-by-step explanation:

Given

v = - \frac{1}{x} YW

Multiply both sides by x to clear the fraction

vx = - YW ( multiply both sides by - 1 )

- vx = YW ( divide both sides by Y )

- \frac{vx}{Y} = W

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Nataly [62]
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5 0
4 years ago
A sequence consists of 20102010 terms. Each term after the first is 11 larger than the previous term. The sum of the 20102010 te
Nataliya [291]

You're considering a sequence of in which consecutive terms differ by 1, meaning

<em>a(n)</em> = <em>a</em> (<em>n</em> - 1) + 1

so <em>a(n)</em> is an arithmetic sequence. (I'm guessing 20102010 should actually be 2010, and 53075307 should be 5307, so 11 should probably be just 1.)

The sum of the first 2010 terms is 5307, or

\displaystyle\sum_{n=1}^{2010}a(n)=5307

Find the value of the first term in the sequence, <em>a</em>(1).

We can write <em>a(n)</em> in terms of <em>a</em>(1) by iterative substitution:

<em>a(n)</em> = <em>a</em>(<em>n</em> - 1) + 1

<em>a(n)</em> = (<em>a</em>(<em>n</em> - 2) + 1) + 1 = <em>a</em>(<em>n</em> - 2) + 2

<em>a(n)</em> = (<em>a</em>(<em>n</em> - 3) + 1) + 2 = <em>a</em>(<em>n</em> - 3) + 3

and so on, down to

<em>a(n)</em> = <em>a</em>(1) + <em>n</em> - 1

So the sum of the first 2010 terms is

\displaystyle\sum_{n=1}^{2010}a(n)=\sum_{n=1}^{2010}\left(a(1)+n-1\right)=(a(1)-1)\sum_{n=1}^{2010}1+\sum_{n=1}^{2010}n=5307

Recall that

\displaystyle\sum_{n=1}^N1=\underbrace{1+1+\cdots+1}_{N\text{ times}}=N

and

\displaystyle\sum_{n=1}^Nn=1+2+\cdots+N=\dfrac{N(N+1)}2

So we have

\displaystyle\sum_{n=1}^{2010}a(n)=2010(a(1)-1)+\frac{2010\cdot2011}2=5307

Solve for <em>a</em>(1) :

2010 (<em>a</em>(1) - 1) + 2,021,055 = 5307

2010 (<em>a</em>(1) - 1) = -2,015,748

<em>a</em>(1) - 1 = - 335,958/335

<em>a</em>(1) = - 335,623/335

Now, every second term, starting with <em>a</em>(1), differs by 2, so they form another arithmetic sequence <em>b(n)</em> given by

<em>b(n)</em> = <em>b</em>(<em>n</em> - 1) + 2

or, using the same method as before,

<em>b(n)</em> = <em>b</em>(1) + 2 (<em>n</em> - 1) = <em>a</em>(1) + 2<em>n</em> - 2

The sum of the 1005 terms in this sequence is

\displaystyle\sum_{n=1}^{1005}b(n)=(a(1)-2)\sum_{n=1}^{1005}1+2\sum_{n=1}^{1005}n

= (- 335,623/335 - 2)•1005 + 2•1005•1006/2

= 1146

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The car can go 32 miles before consuming a full gallon of gas and it traveled 384 miles. So if you divide the 384 miles by 32 miles per gallon, you are left with the gallons consumed.  

 

Once you have this number, you can subtract it from 15 gallons–the total amount of gas in the tank–to get how much gas is left.

Step-by-step explanation:

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Answer:

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