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serg [7]
3 years ago
7

Write an equation in standard form of the line passing through the points (3,3) and (-3,5)

Mathematics
1 answer:
Korvikt [17]3 years ago
8 0

Answer:

The equation of line AB  with points (3,3) and (-3,5) is given as

: x + 3y = 12

Step-by-step explanation:

Here, the given points are A (3, 3) and B (-3,5).

Now, slope of any line is given as :

m = \frac{y_2 - y_1}{x_2 - x_1}

or, m = \frac{5-3}{-3 - 3}   = \frac{2}{-6}  = -\frac{1}{3}

Hence, the slope of the line AB is (-1/3)

Now , A POINT SLOPE FORM of an equation is

(y - y0)  = m (x - x0) ; (x0, y0)  is any arbitrary point on line.

So, for the point (3,3) the equation of the line is

y - 3y-3 = -\frac{1}{3} (x-3)   \implies 3y - 9 = 3 -x

Hence, the equation of line AB  with points (3,3) and (-3,5) is given as:

x + 3y = 12

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Calculate S58 for the arithmetic sequence {an}= {5/6n+1/3} If you could show me the steps that would be great. I keep following
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Answer is 8671/6 which is the third choice

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

Work Shown:

Find the first term of the sequence by plugging in n = 1

a_n = (5/6)*n + 1/3

a_1 = (5/6)*1 + 1/3    replace n with 1

a_1 = 5/6 + 1/3

a_1 = 5/6 + 2/6

a_1 = 7/6

Repeat for n = 58 to get the 58th term

a_n = (5/6)*n + 1/3

a_58 = (5/6)*58 + 1/3    replace n with 58

a_58 = (5/6)*(58/1) + 1/3

a_58 = (5*58)/(6*1) + 1/3

a_58 = 290/6 + 1/3

a_58 = 145/3 + 1/3

a_58 = 146/3

Now we can use the s_n formula below with n = 58

s_n = (n/2)*(a_1 + a_n)

s_58 = (58/2)*(a_1 + a_58)    replace n with 58

s_58 = (58/2)*(7/6 + a_58)   replace a_1 with 7/6

s_58 = (58/2)*(7/6 + 146/3)   replace a_58 with 146/3

s_58 = (58/2)*(7/6 + 292/6)

s_58 = (58/2)*(299/6)

s_58 = (58*299)/(2*6)

s_58 = 17342/12

s_58 = 8671/6


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