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pickupchik [31]
3 years ago
11

24 students in a class took an algebra test. If 18 students passed the test what percent passed?

Mathematics
2 answers:
Leya [2.2K]3 years ago
4 0
18 out of 24 is 18/24 which is 18/24*100=75 percent.
zmey [24]3 years ago
4 0

Answer:75%

Step-by-step explanation:

x% of 24=18

x/100 x 24=18

24x/100=18

Cross multiply

24x=18x100

24x=1800

Divide both sides by 24

24x/24=1800/24

x=75

75 percent pass the test

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Use scalar projection to show that the distance from a point P1(x1,y1) to the line ax+by+c=0 is |ax1+by1+c|/underroot(a^2 +b^2)
Elina [12.6K]

Answer:

d = 13 / 5 units          

Step-by-step explanation:

Given:

- Point P = (-2 , 3 )

- line : 3x - 4y + 5 = 0

Find:

- Use projection to show the given formula

- use the formula to find the distance from given point and line.

Solution:

- Let line L be defined as:

                                     a*x + b*y + c = 0

- Choose A as fixed point on the line L whose coordinates are ( m , k ).

- Now the coordinates of any point (x , y ) can be given by the line L:

                                    a*( x - m ) + b*( y - k ) + c = 0

- We did that by subtracting the coordinates (x,y) from (m,k).

- The above line can represented by dot product of constants (a , b ) to vector coordinates between points (x,y) to (m,k):

                                   ( a , b ) . ( x - m , y - k) = 0

- For above expression to be true i.e the dot product of two vectors is only zero when the vectors are orthogonal. So vector (a , b ) is always perpendicular to every vector in direction of line L

- The vector:

                                         v =  ( x - m , y - k)

- It may represent the line connecting the points (x_1 , y_1) and (m, k).

Then the distance d from the point P_1: (x_1, y_1) to the line L is given by the magnitude of the scalar projection of v onto a, because the latter is the length of the side of a right  triangle with two of the vertices being P_1 and

( m , k), and the other vertex lying on L So:

                                      d = | component v along x |

                                      d =  | v . ( a , b ) | / | (a , b ) |

                                      d = | (x_1 - m , y_1 - k ) . ( a , b)| / sqrt(a^2 + b^2)

                                      d = | ax_1 + by_1 - (a*m + b*k) | / sqrt(a^2 + b^2)

Where point A (m,k) lies on line hence;

                                      a*m + b*k + c = 0

                                      c = - (a*m + b*k)

Hence,

                                      d = | ax_1 + by_1 + c | / sqrt(a^2 + b^2)

- Given point P (-2,3) and line L : 3x - 4y + 5 = 0

where,

                                    (x_1 , y_1) = (-2 , 3 )

                                     (a , b) = ( 3 , -4 )

                                            c = 5

-Evaluate:

                                    d = | 3*(-2) + -4*3 + 5 | / sqrt(3^2 + 4^2)

                                    d = | - 13 | / 5

                                    d = 13 / 5 units                                    

                               

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3 years ago
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