Let the equal sides of the isosceles Δ ABC be x.
Given that the perimeter of Δ ABC = 50m.
Therefore, 2x + AC = 50 --- (1)
It is also given that the perimeter of Δ ABD = 40m.
Therefore, x + BD + AD = 40
BD is the median of the Δ ABC. Therefore, D is the midpoint of AC.
So AD = CD.
Or, AD = 
AC
Therefore, 
Multiply both sides by 2.
2x + 2BD + AC = 80
From (1), 2x + AC = 50.
Therefore, 2BD + 50 = 80
2BD = 80 - 50
2BD = 30
BD = 15m.
Step-by-step explanation:
by using similarity and enlargement
10/8 = 7+y/7
70=56+8y
8y=14
y= 7/4
10/8 = 5+x/5
50=40+8x
x = 8/10 = 4/5
Answer:
The class width is 20
Step-by-step explanation:
In a frequency or a relative frequency distribution the class width is calculated as the difference between the lower or upper class limits of consecutive classes. A point to note is that all the categories or classes usually have the same class width.
We use the first two classes to calculate the class width by using their respective upper limits;
Class width = 89 - 69
Class width = 20
Answer:
B. –7x +7y = –49
Step-by-step explanation:
Graph the line using the slope and y-intercept, or two points.
Slope:
1
y-intercept:
(
0
,
−
7
)
x
y
0
−
7
1
−
6
Answer: n = -6
Step-by-step explanation:
3n+14=-4
Subtract 14 from both sides.
3n+14-14=-4-14
Simplify
3n=-18
Divide both sides by 3
3n/3 = -18/3
n=-6
