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2 years ago

How much tax will be applied to an order for cupcakes that costs $12.75 if the tax rate is 4%?

Mathematics

2 answers:

svetlana [45]2 years ago

7 0

Answer:

$12.75 cupcakes

Multiply the $12.75 x .04 = .51

$12.75 + .51 = $13.26 total

Step-by-step explanation:

klemol [59]2 years ago

4 0

Answer:

$13.26

Step-by-step explanation:

12.75 × 0.04 = 0.51

0.51 + 12.75 = 13.26

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How do I solve for x in this matrix?

Katyanochek1 [597]

When matrices are equal, corresponding terms are equal.

... 5x - 8 = 2

... 5x = 10 . . . . . . . . add 8

... x = 2 . . . . . . . . . divide by 5

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3 years ago

Ajar holding 36 coins, only nickels and quarters, is worth $6.20. There are eight more quarters than there are nickels. How many

Nonamiya [84]

Answer:

Step-by-step explanation:

Let n represent the number of nickels. Then the number of quarters is 36-n. The difference is ...

(36-n) -n = 8 . . . . . 8 more quarters than nickels

18 -n = 4 . . . . . . . divide by 2

14 = n . . . . . . . . add n-4 to both sides

There are 14 nickels in the jar.

_____

<em>Additional comment</em>

The problem supplies more information than is needed for a solution. You can work this as a "sum and difference" problem, as we have above, or you can work it as a "mixture" problem where the total coin value comes into play. Any two of the three given relations will give a solution.

n + q = 36 . . . . . . . number of coins

5n +25q = 620 . . . value of coins (in cents)

q -n = 8 . . . . . . . . difference in number of coins

6 0

2 years ago

There are 33 times as many boys as girls in a room. If 4 boys and 4 girls leave the room, then there will be 5 times as many boy

KATRIN_1 [288]

Answer:

here are 3 times as many boys as girls in a room. If 4 boys and four girls leave the room, then there will be 5 times as many boys as girls in the room. In total, how many boys and girls were in the room originally?

8 girls

24 boys

7 0

2 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

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2 years ago

Please help!!!!!!!!!!

Lina20 [59]

(3+3i) - (13+15i)

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3 years ago