The bill after applying the coupon = 31.84*(85%) = 27.064
so each of them will pay : 27.064/3 = $9.021
Round pan volume is:
3.14•r^2•h
D=7 so r=3.5 in
3.14• (3.5^2)•2 = 76.97 in^3
Rec. pan vol. is :
9•6•2= 108 in^3
Rec. Pan is larger because 108 in^3 is > 76.97 in^3 :) .
The icing that will be needed to frost the round cake pan is:
We need to find the surface area:
S.A= 3.14r^2 + 2 • 3.14•r • h .... 3.14 is the value of PI
So, S.A= 3.14• 3.5^2 + 6.28• 3.5• 2= 82.47 in^2 the icing that'll be needed to frost the round cake pan.
Icing that will be needed for the rec. cake pan is:
2•9•2=36 in^2
6•9•2= 108in^2
6•2= 12 in^2
S.A= 156 in^2 the icing needed to frost the rec. cake pan .... the S.A of all sides except the bottom one :).
Good luck ;-)
how many 3 element subsets of {1, 2, 3, 4, 5, 6, 7, 8, ,9, 10, 11} are there for which the sum of the elements in the subset is
AURORKA [14]
Answer:
There are 155 ways in which these elements casn occur.
Step-by-step explanation:
We want 3 element subsets whose sum are multiples of 3
1+2+3= 6
1+2+6= 9
1+2+9= 12
1+9+11=21
1+3+5=9
1+4+8=12
1+5+6=12
1+6+8=15
1+7+10=18
1+8+9=18
1+9+11=21
2+3+7=12
2+4+6=12
2+4+9=15
2+5+11=18
2+6+7=15
2+7+9=18
2+8+5=15
2+8+11=21
2+9+10=21
3+6+9= 18
3+9+11=21
3+10+11=24
6+9+10=27
6+8+11=27
6+7+11=24
7+8+9= 24
8+9+10=27
7+9+11=27 .........
We have 11 elements
We need a combination of 3
The combinations can be in the form
even+ even+ odd
odd+odd+odd
even + odd+odd
So there are 3 ways in which these elements can occur
Total number of combinations with 3 elements =11C3= 165
There are 6 odd numbers and 5 even numbers.
Number of subsets with 3 odd numbers = 6C3= 20
Number of two even numbers and 1 odd number = 5C2*6C1=10*6= 60
Number of 2 odd and 1 even number = 6C2* 5C1= 5*15= 75
So 20+60+75=155
There are 155 ways in which this combination can occur
Answer:
Step-by-step explanation:
Start by factoring out a 5:
We need to find two integers that have a product of 12, and a sum of -7:
(-3)(-4)=12
-3-4=-7
We can split -7x into -3x and -4x
Factor each half separately:
![5[x(x-3)-4(x-3)]](https://tex.z-dn.net/?f=5%5Bx%28x-3%29-4%28x-3%29%5D)
Since x and -4 are both being multiplied by x-3, we can combine them:
