Answer: none
Step-by-step explanation:
(A)
(16÷32/10) ×2 + 0.2×(90)
Using bodmas principle ; solve bracket
(16×10/32)×2 + (2/10×90)
10+18 =28
(B)
{(16÷32/10) × (2+2/10)} ×90
Open brackets
{(16×10/32) × (22/10)} ×90
(5×11/5) ×90
11×90 = 990
(C)
16÷{(32/10×2) + (2/10×8)} +82
Open brackets, solve division first, dolled by addition
16÷(32/5 + 8/5) +82
16÷(40/5) +82
16÷8 +82
2+82= 84
(D)
[16÷(32/10 ×2) + 0.2× (90)]
16÷ (32/5) + 2/10 ×90
Solve division
16×5/32 + 18
5/2 + 18
L.c.m of denominator (2&1) =2
(5+36) / 2 = 41/2
=20.5
20l + 25g -10
replace G with 4 and L with 3
20(3) + 25(4) - 10
60 + 100 - 10
160 - 10
150
he made $150
Let
x--------> the amount of gummy candy in pounds
y--------> the amount of jelly beans in pounds
z--------> the amount of hard candy in pounds
we know that
--------> equation 
--------> equation 
--------> equation 
Substitute equation in equation and equation
![[2y]+y+z=9](https://tex.z-dn.net/?f=%5B2y%5D%2By%2Bz%3D9)
-----> 
------> equation 
![2[2y]+3y+3z=23](https://tex.z-dn.net/?f=2%5B2y%5D%2B3y%2B3z%3D23)
-----> 
------> equation 
using a graphing tool ------> Solve the system of equations
see the attached figure
the solution is the point 
<u>Find the value of x</u>
therefore
<u>the answer is</u>
the amount of gummy candy is 
the amount of jelly beans is 
the amount of hard candy is 
Answer:
Step-by-step explanation:
10 contain both chocolate and caramel
There are 18 candies.
3 don't contain either chocolate or caramel
12 contain chocolate.
There are 18 - 12 - 3 that are not accounted for. So we have 3 that are not mentioned. Those 3 must contain just caramel.
So it should be filled out like this
contain caramel Do not contain Caramel
10 2
3 3
