Answer:
A. 40x + 10y + 10z = $160
B. 8 Roses, 2 lilies and 2 irises
C.
1. 20x + 5y + 5z = $80
2. 4x + y + z = $16
3. 8x + 2y + 2z = $32
Step-by-step explanation:
Cost for each flower = $160/5 = $32
So we have $32 for each bouquet consisting of 12 flowers each.
Roses = x = $2.50 each
lilies = y = $4 each
irises = z = $2 each
8x + 2y + 2z = $32
8($2.50) + 2($4) + 2($2) = $32
$20 + $8 + $4 = $32
$32 = $32
a. Maximum budget is $160
40x + 10y + 10z = $160
40($2.50) + 10($4) + 10($2) = $160
$100 + $40 + $20 = $160
$160 = $160
b. From above
8x + 2y + 2z = $32
8 Roses, 2 lilies and 2 irises
c. No. There are other solutions If total cost is not limited
1. 20x + 5y + 5z
20($2.50) + 5($4) + 5($2)
$50 + $20 + $10
= $80
2. 4x + y + z
4($2.50) + $4 + $2
$10 + $4 + $2
= $16
3. 8x + 2y + 2z
8($2.50) + 2($4) + 2($2)
$20 + $8 + $4
= $32
Answer:
circle= 5, triangle= 3
Step-by-step explanation:
Please see attached picture for full solution.
The answer is digit sum method.
Digit sum method is method used to check the sum of sum numbers. If the sum of all of the digits of numbers is equal to the sum of all of the digits of the total sum, then the arithmetic process was correct.
We need to check the sum of <span>104+34+228+877:
</span>104 + 34 + 228 + 877 = 1243
Let's first summarize the digits of individual numbers:
104 *** 1 + 0 + 4 = 5
34 *** 3 + 4 = 7
228 *** 2 + 2 + 8 = 12 *** 1 + 2 = 3
877 *** 8 + 7 + 7 = 22 *** 2 + 2 = 4
Now, let's sum these sums:
5 + 7 + 3 + 4 = 19 *** 1 + 9 = 10 *** 1 + 0 = <u><em>1</em></u>
Then, let's summarize the digits of the total sum:
1243 *** 1 + 2 + 4 + 3 = 10 *** 1 + 0 = <u><em>1</em></u>
Since the sums of the digits on the both sides of equation is 1, than the arithmetic process was correct and the sum of <span>104 + 34 + 228 + 877 is really 1243.</span>
Answer: 210
<em>Step-by-step explanation: 70+70+70=210 and 40% for 210 students would add up to 120% of all students surveyed. Look if there is 70 students that's been surveyed that is 40% add 40% 3 times(40%+40%+40%)=120% of all students surveyed </em>
