The value of "x" is 5.

Given that,
A trapezium ABCD in which AB || CD such that
Now,











Answer:
a?
Step-by-step explanation:
Answer:
12
Step-by-step explanation:
Givens
Harpreet = x + 3
Sukhpreet = x
Komalpreet = 2*(x + 3)
Average Age = 15
Equation
[(x + 3) + x + 2*(x + 3) ] / 3 = 15 Multiply both sides by 3
Solution
3 [(x + 3) + x + 2*(x + 3) ] / 3 = 15 * 3 Combine
[(x + 3) + x + 2*(x + 3) ] = 45 Remove the brackets
x + 3 + x + 2x + 6 = 45 Combine like terms
4x + 9 = 45 Subtract 9 from both sides
4x + 9 - 9 = 45 - 9 Combine
4x = 36 Divide by 4
4x/4 = 36/4
x = 9
Answer
Harpreet is x+ 3 = 9 + 3 = 12
Answer:

Step-by-step explanation:
Use the slope-intercept form to write the equation if the line of the graph given.
, where,
b = y-intercept, which is where the line cuts across the y-axis = 4.
m = slope = 
Substitute m = ⁴/3 and b = 4, into the slope-intercept formula to get the equation of the line:


Answer:
168 in², 94.25 yd²
Step-by-step explanation:
I couldn't find a formula for the SA of a triangular prism so I just found the area of each surface.
The formula for SA of a cylinder is 
All work for triangular prism is in the image