To solve this problem, we must imagine the triangles and
parallel lines which are formed. It is best to draw the triangle described in
the problem so that you can clearly understand what I will be talking about.
The first step we have to do is to make an equality equation
in triangle ABC.
In triangle ABC, we are given that lines XY and BC are two
parallel lines (XY || BC). Therefore
this means that:
AX / XB = AY / YC --->
1
The next step is to make an equality equation in triangle
AXC.
We are given that lines ZY and XC are two parallel lines (ZY
|| XC). Therefore this also means that:
AZ / ZX = AY / YC ---> 2
Combining 1 and 2 since they have both AY / YC in common:
AX / XB = AZ / ZX
we are given that:
AZ = 8, ZX = 4 therefore AX = AZ + ZX = 12, hence
12 / XB = 8 / 4
XB = 6
- The domain of the attachment graph .
The method to identify the domain and range of functions is by using graphs :-
⚡ The domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis .
⚡ The range is the set of possible output values, which are shown on the y-axis .
See the attachment figure .
[NOTE :- Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values .]
✍️ So, we can observe that the graph extends horizontally from -3 to the right with +10 .
✯ Hence, the domain is [-3 , 10] .
Step-by-step explanation:
x +2y = 9. .....(1)
x - 2y = 5. .....(2)
from eqn (1)
x + 2y = 9
x = 9 - 2y.
substitute x = 9 - 2y into eqn (2)
9 - 2y - 2y = 5
-4y = 5 - 9
-4y = -4
y = 1
sub y = 1 into eqn (1)
x + 2y = 9
x + 2(1) = 9
x + 2 = 9
x = 9 - 2
x = 7.
(2) x + y = 9 ....(1)
x - 2y = 0 .....(2)
from eqn (2)
x - 2y = 0
x =0 + 2y
substitute x = 2y into eqn (1)
x + y = 9
(2y) + y = 9
3y = 9
y = 3
substitute y = 3 into eqn (2)
x - 2y = 0
x - 2(3) = 0
x - 6 = 0
x = 6.
(3) 2x + 7y = 5. ....(1)
2x + 3y = 9. .....(2)
from eqn (1)
2x + 7y = 5
7y = 5 - 2x
y = (5 - 2x)/7
sub y = (5 - 2x)/7 into eqn (2)
2x + 3y = 9
2x + 3(5 - 2x)/7 = 9
2x + (15 - 6x)/7 = 9
multiply through by 7
14x + 15 - 6x = 63
14x - 6x = 63 - 15
8x = 48
x = 6
sub x = 6 into eqn (1)
2x + 7y = 5
2(6) + 7y = 5
12 + 7y = 5
7y = 5 - 12
7y = -7
y = -1
The answer is 2/3 x 18= 12.
