6(2k-3) is the answer to this problem.
make a distributive
take the 6 out of the equation and u have remaining 2k-3. 6(2k-3)
4t=r
a=pir^2
sub 4t for r
a=pi(4t)^2
a=pi16t^2
a(t)=16pi(t^2)
A. a(t)=16pi(t^2)
B. sub 4 for t
a(4)=16pi4^2
a(4)=16pi16
a(4)=16*16*3.14
a(4)=803.84 square units
A. a(t)=16pi(t^2)
B. 803.84 square units
For the first digit you are choosing from 2 digits
for the second digit you are choosing from 2 digits
for the third digit you are choosing from 2 digits
2*2*2=8
111
110
011
101
000
001
100
010
However......
A number doesn't usually start with a 0 or 0s.
Therefore, if you want 3-digit numbers and not just permutations using 0 and 1, then you must eliminate
011, 000, 001, and 010
Seeing that the first digit can't be 0, you choose from 1, then 2, and then 2 digits again; 1*2*2=4 numbers
You choose which answer best suits your problem.
He rowed 10 miles in total. YOU'RE WELCOME :D
Answer:
<h2>22</h2>
Step-by-step explanation:
<h3>Given, y = 3 </h3>
substitute this value in given expression
<em>=> 10y - 8</em>
<em>=</em><em>></em><em> </em><em>1</em><em>0</em><em>(</em><em>3</em><em>)</em><em> </em><em>-</em><em> </em><em>8</em><em> </em>
<em>=</em><em>></em><em> </em><em>3</em><em>0</em><em> </em><em>-</em><em> </em><em>8</em><em> </em>
<em>=</em><em>></em><em> </em><em>2</em><em>2</em><em>.</em><em>.</em><em>.</em><em>ans</em>
<h2>HOPE IT HELPS U!!!!</h2>
