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jasenka [17]
2 years ago
5

Write an equation of the line with the same slope as the line given by 5x + 2y = 8 and the same y-intercept as the line given by

3x - 7y = 10
Mathematics
1 answer:
kaheart [24]2 years ago
8 0
In y = mx + b form, the slope is in the m position and the y intercept is in the b position.

5x + 2y = 8
2y = -5x + 8
y = -5/2x + 8.....slope (m)  is -5/2

3x - 7y = 10
-7y = -3x + 10
y = 3/7x - 10/7.....y int (b) = -10/7

so ur equation with slope(m) of - 5/2 and y int (b) of -10/7 is :
y = -5/2x - 10/7 <== slope intercept form



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Step-by-step explanation:

36-24=12 and 12/2 is 6

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How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used
MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by

• <em>a</em> (1) = 1

• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1

We can find the explicit rule for the sequence by iterative substitution:

<em>a</em> (2) = <em>a</em> (1) + 21

<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21

<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21

and so on, with the general pattern

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20

Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is

<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332

or more compactly,

\displaystyle\sum_{n=1}^N a(n) = 2332

It's important to note that <em>N</em> must be some positive integer.

Replace <em>a</em> (<em>n</em>) by the explicit rule:

\displaystyle\sum_{n=1}^N (21n-20) = 2332

Expand the sum on the left as

\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332

and recall the formulas,

\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n

\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2

So the sum of the first <em>N</em> terms of <em>a</em> (<em>n</em>) is such that

21 × <em>N</em> (<em>N</em> + 1)/2 - 20<em>N</em> = 2332

Solve for <em>N</em> :

21 (<em>N</em> ² + <em>N</em>) - 40<em>N</em> = 4664

21 <em>N</em> ² - 19 <em>N</em> - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.

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