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How much is 3 dozen cakes at 45c each?

Updated: 4/28/2022
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Wiki User

12y ago

Best Answer

.45*(3*12)= $16.20

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12y ago
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AMELIA KENNEY

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2y ago

$16.00

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Kylee Homenick

Lvl 1
2y ago
how did you get that answer?
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Al Leuschke

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2y ago
great answer thxxx

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Anonymous

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3y ago

16.20

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Anonymous

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3y ago

8

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Q: How much is 3 dozen cakes at 45c each?
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Continue Learning about Other Math

How do you solve an equation with a variable on each side?

It's not always to 'solve' as it is to 'simplify' EX: 15a + 3b = 45c simplifies to 5a + b = 15c


What would this mean (In a maths state) 45c/5c?

9


What are math expressions?

A math expression is basically and equation without and equals sign. For example "15a+30b" would be an expression, while "15a+30b=45c" would be an equation. An expression can also be defined as a sum of terms.it could also be defined as a mathematical phrase which involves or contain variables and numbers.like(x and y)these are variables.


A piggy bank has only nickels and dimes in it there are 14 coins worth 0.95. How many nickels and dimes are in the bank?

2 variables with 2 equations. A simultaneous equation problem. d+ n = 14 10d + 5n = 95 Multiply the first equation by -5 -5d + -5n = -70 10d + 5n = 95 add together 5d = 25 d = 5 ◄ plug that in the first equation to find n 5 + n = 14 n = 9 ◄ check 5 dimes plus 9 nickles is 50c+45c=95c ■


What number when divided by 3 leave a remainder of 1 divided by 4 leaves a remainder of 2 divided by 5 leaves a remainder of 3 and divided by 6 leaves a remainder of 4?

This is actually a little bit tricky to work, without just trying numbers. The first answer is 58, however there are an infinite amount of correct answers. For example 118, 178, and 238 are the next correct answers in sequence. In fact, the general solution to this answer is:n = 58 + 60k, where k is just any positive integer (0, 1, 2, 3, 4, 5, ...)In the equation n = 58 + 60k, the 60 comes from the fact that it is the least common multiple of 3, 4, 5, and 6. The 58 is a little harder to explain. Here is how to algebraic solve this. It involves the modulus operation, so you should be familiar with that before reading the steps on how to systematically solve this problem.n = dk + r, where n is the answer, d is the number we are dividing by, k is some integer, and r is the remainder.k will vary from each condition, so let a,b,c,d = some integern = 3a + 1n = 4b + 2n = 5c + 3n = 6d + 4If we rewrite these as a modulus we have (1 below corresponds to 1 above):n%3 = 1 (% = modulus)n%4 = 2n%5 = 3n%6 = 4Now take one of the n terms and place it into a different modulus rewrite:We'll use the first term (#1) and the second term (#2).n = 3a + 1 AND n%4 = 2( 3a + 1 ) % 4 = 2For simplicity, let's make the remainder zero, instead of two.( 3a + 1 - 2 ) % 4 = 2 - 23a - 1 % 4 = 0We need the left hand side to be some multiple of 4 for this equation to hold true. If a = -1 (find -1 by trial and error)3(-1) - 1 = -3 - 1 = -4, and -4%4=0 is true, so this value of a worksLet j = some integera = -1 + 4j (we can add any multiple of 4 to our solution)Now we need to do the exact same thing with the second two conditions (terms #3 and terms #4):n = 5c + 3 AND n%6 = 45c + 3 % 6 = 45c - 1 % 6 = 0Once again, if we try c = -1, we end up with:5(-1) - 1 % 6 = 0-5 - 1 % 6 = 0-6%6=0 which holds trueLet k be some integerTherefore, c = -1 + 6kPlug a back into it's orginial equation:n = 3a + 1, where a = -1 + 4jn = 3(-1 + 4j) + 1 = 12j - 2And Plug c back into it's originial equation:n = 5c + 3, where c = -1 + 6kn = 5(-1 + 6k) + 3 = 30k - 2Now we have:n = 12j - 2n = 30k - 2We have to do the same process as above with the new terms:n = 12j - 2 is the same as saying n%12=-2However, that doesn't make since, so add a multiple of 12 to itn%12=-2(+12)=10We only have to convert on term to the modulus form, and plug the other form into it, just like before:n%12=10 AND n = 30k - 230k-2%12=1030k-12%12=0because 12 is just a multiple of itself, 12, we can drop the -12 (if you don't drop it you will end up with the same answer)30k%12=0Simplify your answer (divide both terms by 6):5k%2=0We need to find a value of k, where 5k%2=0 holds true:if k=2: 10%2=0 (which is true)Let x = some integerSo k=2xPlug k back into an equation:n = 30k - 2, where k=2xn = 30(2x) - 2 = 60x - 2Because we have now included every condition, and reduced our equation to only one independent variable, we have our answer:n = 60x - 2, where x is some integerif x=0: n=-2 (which all conditions hold true on)if x=1: n=58 (which is the answer I provided up top)if x=2: n=118if x=3: n=178if x=4: n=238and so on and so forthn = 60x - 2 is the same as the equation I provided above, it's just offset by 60.