First make 20 into a fraction, then solve; 3/5 · 20/1. 60/5 = 12. Hope that helps!
So polygon B measures 20 and polygon A is four times smaller than you would divide that 20 by for and get 5 for your answer:)
The larger number is 13 and smaller number is 4
<em><u>Solution:</u></em>
Let the larger number be "a"
Let the smaller number be "b"
<em><u>Given that larger of two numbers is twice the smaller increased by five</u></em>
larger number = twice the smaller increased by five
larger number = twice the smaller number + 5
a = 2b + 5 --- eqn 1
<em><u>Also given that three times the larger exceeds double the smaller by 31</u></em>
Here "exceeds" denotes addition and "times" denotes multiplication
"double" denotes being multiplied by 2
three times the larger number = double the smaller number + 31
3a = 2b + 31 ---- eqn 2
<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "b"</u></em>
Substitute eqn 1 in eqn 2
3(2b + 5) = 2b + 31
6b + 15 = 2b + 31
6b - 2b = 31 - 15
4b = 16
<h3>b = 4</h3>
From eqn 1,
a = 2b + 5
a = 2(4) + 5
a = 8 + 5
<h3>a = 13</h3>
Thus the larger number is 13 and smaller number is 4
I would say a line graph is probably the best way (it is usually the normal way) so that you can see the fluctations in temperature. Time on the x-axis and temp on the y-axis
