There could only be 8 because 10 *8 is 80 and that is your maximum number you can have
Hi there!
PART A:
The system of equations we would use would be:
x + y = 22 (amount of items)
3x + 1y = 30 (cost)
Variables:
x = the amount of strawberry wafers bought at the price of $3
y = the amount of chocolate wafers bought at the price of $1
PART B:
To solve, we'll use substitution because we can easily isolate a variable using the first equation.
Work:
x + y = 22 (first equation)
y = 22 - x (isolating a variable)
3x + 1y = 30 (second equation)
3x + (22 - x) = 30 (substituting into the second equation)
2x + 22 = 30 (simplifying)
2x = 8 (subtracting)
x = 4 strawberry wafers
4 + y = 22 (substituting x into the first equation to solve for y)
y = 18 chocolate wafers
ANSWER:
They bought 4 strawberry wafers and 18 chocolate wafers.
Hope this helps!! :)
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Answer:
a) 375
b) 7062.75 mm²
Step-by-step explanation:
b) We need to find the shortest possible width and length to get the smallest possible area.
To get the boundaries for 19.4, we go on to the next significant figure (the hundredths) and ± 5 of them.
The boundaries are, therefore: 19.35 - 19.45
As for the length, we can see they've added 5 units as the measurement is correct to 2 sig' figures, which is the tens.
And so, if we do as we did before, we go to the next sig' figure (the units) and ± 5 of them, we get the boundaries to be 365 - 375.
Now, we just multiply the lower bounds of the length and width to get the minimal/lower-bound area:
365 * 19.35 = 7062.75 mm²
T is a linear function of C ==> T = mC + b
Each additional chirp C corresponds to an increase in T of 0.25 ==> m = 0.25
T = 52 when C = 20 ==> 52 = 0.25(20) + b ==> b = 52 - 5 = 47
So T = 0.25C + 47
When C = 100 then T = 0.25(100) + 47 = 25 + 47 = 72
When C = 120 then T = 0.25(120) + 47 = 30 + 47 = 77 so the
temperature range is between 72 and 77 when there are between 100 and
120 chirps per minute
The greatest area he can fence is 64 ft².
In order to maximize area and minimize perimeter, we use dimensions that are as close to equivalent as possible. 32 feet of fence for 4 sides gives us 8 feet of fence per side. We would have a square whose side length is 8; the area would be 8*8 = 64.
