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nydimaria [60]
3 years ago
7

Some of the students three scores is 231. If the first is 20 points more than the second, and the sum of the first two is 6 more

times the third, what was the first score?
Mathematics
1 answer:
Aleks [24]3 years ago
4 0

Answer:

The first score is 109

Step-by-step explanation:

I am assuming that in the first sentence of the question, you meant:

Sum of the students three scores is 231...

First, let the scores of the first second and third student be a, b and c respectively. We are told that:

a + b + c = 231 . . . . . . . .(1)         (sum of students three scores is 231)

a = b + 20 . . . . . . . . . . . (2)        (the first is 20 points more than the second)

a + b = 6c . . . . . . . . . . . .(3)        (sum of the first two is 6 more times the third)

required, find a.

substituting the value of (a + b) in equation (3) into equation (1), we will have the following:

since a + b = 6c . . . (3)

a + b + c = 231 . . . . . (1), becomes,

(a + b) + c = 231

(6c)  + c = 231

7c = 231 (divide both sides by 7)

c = 231 ÷ 7 = 33

∴ c = 33

Next, from equation (2), we know that a = b + 20; this can also be written as:

a - 20 = b

∴ b = a - 20 . . . . . . . (4)

Finally, putting the value of b in equation (4) and the value of c calculated above into equation 1, ( a + b + c = 231), we have the following:

a + (a - 20) + 33 = 231

(a + a) - 20 + 33 = 231

2a + 13 = 231

2a = 231 - 13 = 218

a = 218 ÷ 2 = 109

∴ a = 109

we can also calculate for 'b' by substituting for the value of 'a' in equation 4

b = a - 20 = 109 - 20 = 89.

and to test if the values of a, b and c are correct:

a + b + c = 231

109 + 89 + 33 = 231

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\bold{\huge{\underline{ Solution }}}

<u>We </u><u>have</u><u>, </u>

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Let the coordinates of the end point of the line segment AB be ( x1 , y1 ) and (x2 , y2)

<u>Also</u><u>, </u>

Let the coordinates of midpoint of the line segment AB be ( x, y)

<u>We </u><u>know </u><u>that</u><u>, </u>

For finding the midpoints of line segment we use formula :-

\bold{\purple{ M( x,  y) = }}{\bold{\purple{\dfrac{(x1 +x2)}{2}}}}{\bold{\purple{,}}}{\bold{\purple{\dfrac{(y1 + y2)}{2}}}}

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

  • The coordinates of midpoint and one of the end point of line segment AB are ( -8,8) and (-2,-20) .

<u>For </u><u>x </u><u>coordinates </u><u>:</u><u>-</u>

\sf{  -8  = }{\sf{\dfrac{(- 2 +x2)}{2}}}

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\sf{ - 16 = - 2 + x2 }

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\bold{ x2 = -14  }

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\sf{  8  = }{\sf{\dfrac{(- 20 +x2)}{2}}}

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The domain of f/g consists of numbers x for which g(x) cannot equal 0 that are in the domains of both f and g.

 

Let’s take this equation as an example:

If f(x) = 3x - 5 and g(x) = square root of x-5, what is the domain of (f/g)x. 


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x must be in the domain of f: f(x) = 3x -5 are in the domain of x and all real numbers x.

x must be in the domain of g: g(x) = √(x - 5) so x - 5 ≥ 0 so x ≥ 5.

g(x) can not be 0: g(x) = √(x - 5) and √(x - 5) = 0 gives x = 5 so x ≠ 5.

 

Hence to x x ≥ 5 and x ≠ 5 so the domain of (f/g)(x) is all x satisfying x > 5.

 

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Now that we have the parenthesis taken care of we can do the simpler math,

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I just added the 15 and 8, so now I move the -5.5 to the opposite side by adding.

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Here we can do the same with the -2m.

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Sorry it's really long, but I hope this helps! Have a great day!

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