Answer:
I will try to answer this shortly it's tricky
So even postive integers are by defention in form 2k where k is a natural number so
let the sum of even integers to n=S
S=2(1+2+3+4+5+6+7+8+......+k-1+k
divide bith sides of equation 1 by 2
0.5S=1+2+3+4+5+...........+k-1+k
S=2(k+(k-1)+..............................+2+1)
divide both sides of equation 2 by 2
0.5S=k+k-1+..............................+2+1)
by adding both we will get
___________________________
S=(k+1)(k)
so the sum will be equal to
S=

so let us test the equation
for the first 3 even number there sums will be
2+4+6=12
by our equation 3^2+3=12
gave us the same answer so our equation is correct
Answer:
I am not completely sure but I believe that it is C) 240 yd^3
Step-by-step explanation:
Answer:
<h2><em><u>120</u></em><em><u> </u></em><em><u>square</u></em><em><u> </u></em><em><u>units</u></em><em><u> </u></em></h2>
Step-by-step explanation:
<em><u>Given</u></em><em><u>, </u></em>
Base of the parallelogram = 10
Height of the parallelogram = 12
<em><u>Therefore</u></em><em><u>, </u></em>
Area of the parallelogram = <em>base</em><em> </em><em>×</em><em> </em><em>height</em><em> </em>
= 10 × 12
= 120
<em><u>Hence</u></em><em><u>,</u></em>
<em><u>Area</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>parallelogram</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>120</u></em><em><u> </u></em><em><u>square</u></em><em><u> </u></em><em><u>units</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em>
Answer:
Z.
Step-by-step explanation:
I wrote down the subject with the number of students. I looked one by one to see if the number of students with the subject matched with the graph. Then I finally found that the last graph matched with what I written down.