
is a right triangle with base length 1 and height 8, so the area of

is

.
The average value of

over

is given by the ratio

The denominator is just the area of

, which we already know. The average value is then simplified to

In the

-plane, we can describe the region

as all points

that lie between the lines

and

(the lines which coincide with the triangle's base and hypotenuse, respectively), taking

. So, the integral is given by, and evaluates to,



Answer: The area of the triangle is 252 :)
Step-by-step explanation:
M = 6
L = 6
K = 7
6 x 6 = 36 x 7 = 252
The area of the triangle is 252 :)
Answer:
B) 70.5
Step-by-step explanation:
Median = middle value when the values are placed in order.
If there are two middle values, the median is the mean of those two values.
<u>Class 1</u>
45 46 51 52 53 53 61 63 64 65 66 68 70 <u>70 71</u> 73 76 77 79 81 82 83 84 87 90 92 93 95
There are 28 values in Class 1.
Therefore, there are two middle values: 14th and 15th values
14th value = 70
15th value = 71

Factoring may be the easiest one
Answer:

Step-by-step explanation:


We have to find


If 

=
Hence our answer is
