132, to sum up, the Lcm Of 11 and 12 is 132
Answer:
Q1. x= 18, y=59
Q2. m∠J= 56°
Step-by-step explanation:
Q1. (3x +5)°= y° (base ∠s of isos. △)
y= 3x +5 -----(1)
(3x +5)° +y° +(4x -10)°= 180° (∠ sum of △)
3x +5 +y +4x -10= 180
7x +y -5= 180
7x +y= 180 +5
7x +y= 185 -----(2)
Substitute (1) into (2):
7x +3x +5= 185
10x= 185 -5
10x= 180
x= 180 ÷10
x= 18
Substitute x= 18 into (1):
y= 3(18) +5
y= 59
Q2. (5x -13)°= (3x +17)° (base ∠s of isos. △)
5x -13= 3x +17
5x -3x= 17 +13
2x= 30
x= 30 ÷2
x= 15
∠LKJ
= 3(15) +17
= 62°
∠KLJ= 62° (base ∠s of isos. △)
m∠J
= 180° -62° -62° (∠ sum of △JKL)
= 56°
Answer:
2 2/19
Step-by-step explanation:
5 cases in 2 3/8 hours
The unit rate of cases per hour is calculated by dividing the number of cases by the number of hours.
5/(2 3/8) = 5/(19/8) = 5 * 8/19 = 40/19 = 2 2/19
3x2x(-4)=-24
3x2-4=2
3+2-4=1
not entirely sure what the equation your asking is but i hope this helps
The distance between starting and ending point is 34 miles.
Step-by-step explanation:
Given,
Car moves 16 miles to north then 30 mile to east.
It forms a right angle triangle.
The straight line distance from starting to ending point represents hypotenuse.
To find the distance between starting and ending point.
Formula
By <em>Pythagoras theorem,</em>
h² = b²+l² where h is the hypotenuse, b is base and l is the another side.
Taking, b=16 and l=30 we get,
h² = 16²+30²
or, h = 
or, h = 
= 34
Hence,
The distance between starting and ending point is 34 miles.
