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Aleksandr-060686 [28]
3 years ago
5

How do u figure out the rise and run without a graph

Mathematics
1 answer:
Sphinxa [80]3 years ago
6 0
Change in y over change in x aka y2-y1/x2-x1
Plug in the x and y variables and solve.
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Find the equation of a line, in slope intercept form of a line that passes through the point (9,2) and is perpendicular to the l
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3 years ago
(-2,1); perpendicular to y= - 2/5x -4<br><br> Help explain step by step?
Fittoniya [83]

Answer:

  y = (5/2)x + 6

Step-by-step explanation:

The given line has a slope (coefficient of x) of -2/5. The perpendicular line will have a slope that is the negative reciprocal of this:

  -1/(-2/5) = 5/2

Since you are given a point you want the line to go through, the point-slope form of the equation of the line is useful. That form for slope m and point (h, k) is ...

  y = m(x -h) +k

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  y = 5/2(x +2) +1

  y = 5/2x +6 . . . . . eliminate parentheses

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4 years ago
Rowena can paint a room in $14$ hours, while Ruby can paint it in $6$ hours. If Rowena paints for $x$ hours and Ruby paints for
LUCKY_DIMON [66]

Question

Rowena can paint a room in 14 hours, while Ruby can paint it in 6 hours. If Rowena paints for x hours and Ruby paints for y hours, they will finish half of the painting, while if Rowena paints for y hours and Ruby paints for x hours they will paint the whole room. Find the ordered pair (x,y)

Answer:

(x, y) = (231/40, 21/40)

where x = 231/40

y = 21/40

Step-by-step explanation:

From the question, we are told:

Rowena can paint a room in = 14 hours

Ruby can paint it in = 6 hours.

This means

In one hour.

Rowena can paint = 1/14 of the room

Ruby can paint =1/6 hour of the room

From the question, we are told that :

If Rowena paints for x hours and Ruby paints for y hours, they will finish half of the painting

This is represented mathematically as:

(1/14)(x) + ( 1/6)(y) = 1/2...... Equation 1

Also we were told that:

if Rowena paints for y hours and Ruby paints for x hours they will paint the whole room

(1/14)(y) + (1/6)(x) = 1 .......... Equation 2

Bringing the two equations together, we have:

(1/14)(x) + ( 1/6)(y) = 1/2 ....... Equation 1

(1/14)(y) + (1/6)(x) = 1 ............ Equation 2

We find the Lowest common multiple of the numerator 14 and 6 = 42. Hence, we multiply both equations through by 42

3x + 7y = 21 ....... Equation 3

7x + 3y = 42 ......... Equation 4

To solve for x and y in Equation 3 and 4 we would be using the Elimination method

21x + 49y = 147 .......... Equation 5

21x +9y = 126 ........... Equation 6

21x + 49y = 147 .......... Equation 5

-(21x +9y = 126) ........... Equation 6

40y = 21

y = 21/40

To get the value of x , we would substitute 21/40 for y in Equation 3

3x + 7y = 21 ....... Equation 3

3x + 7(21/40) = 21

3x + 147/40 = 21

3x = 21 - 147/40

3x = 693/40

x = (693/40) ÷ 3

x = 693/40 × 1/3

x = 693/120

x = 231/40

Therefore, (x, y) = (231/40, 21/40)

3 0
3 years ago
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