You can solve this either just plain algebra or with the use of trigonometry.
In this case, we'll just use algebra.
So, if we let M be the the point that partitions the segment into a ratio of 3:2, we have this relation:
KM/ML = 3/2
KM = 1.5 ML
We also have this:
KL = KM + ML
Substituting KM,
KL = (3/2) ML + ML
KL = 2.5 ML
Using the distance formula and the given coordinates of the K and L, we get the length of KL
KL = sqrt ( (5-(-5)^2 + (1-(-4))^2 ) = 5 sqrt(5)
Since,
KL = 2.5 ML
Substituting KL,
ML = (1/2.5) KL = (1/2.5) 5 sqrt(5) = 2 sqrt(5)
Using again the distance formula from M to L and letting (x,y) as the coordinates of the point M
ML = 2 sqrt(5) = sqrt ( (5-x)^2 + (1-y)^2 ) [let this be equation 1]
In order to solve this, we need to find an expression of y in terms of x. We can use the equation of the line KL.
The slope m is:
m = (1-(-4))/(5-(-5) = 0.5
Using the general form of the linear equation:
y = mx +b
We substitue m and the coordinate of K or L. We'll just use K.
-5 = (0.5)(-4) + b
b = -1.5
So equation of the line is
y = 0.5x - 1.5 [let this be equation 2]
Substitute equation 2 to equation 1 and solving for x, we get 2 values of x,
x=1, x=9
Since 9 does not make sense (it does not lie on the line), we choose x=1.
Using the equation of the line, we get y which is -1.
So, we get the coordinates of point M which is (1,-1)
All the steps were correct except the final statement. The
mistake was in Line 6.
Line 6 triangle ABC is congruent to triangle EFD by
SAS.
<span>This does not follow. The SAS postulate states
that if two sides and the included angle of one triangle is congruent to two sides
and the included angle of another triangle. The student only proved that one side
of the triangle (AC) is congruent to the side of another triangle (EF) .</span>
Answer:B is the answer -11
Step-by-step explanation:
Let the number of chemistry books be x and the number of calculus books be y
total number of books in each case is:
x+y=24
x=24-y.....i
Weight of each case:
2707.2/20
=135.36
thus
4.2x+6.0y=135.36..ii substituting i in ii
4.2(24-y)+6y=135.36
100.8-4.2y+6y=135.36
hence
1.8y=34.56
y=19.2
hence
x=24-19.2
x=4.8
thus the number of chemistry books was 4.8*20=96 number of calculus was
19.2*20
=384 books
