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solmaris [256]
2 years ago
9

5 more than the quotient of a number and 8 is 42

Mathematics
2 answers:
Zepler [3.9K]2 years ago
3 0

Answer:

<u>The number is 296.</u>

Step-by-step explanation:

Let's call "x" to the number we are looking for.

Now, the problem states that "5 more than the quotient of a number and 8 is 42". This means that this number is being divided by 8, it's also being additioned 5 and the final result is 42. Therefore, the expression of this operations on this number is the following:

\frac{x}{8}+5=42. Let's solve the equation to find x.

1. Write the expression.

\frac{x}{8}+5=42

2. Substract 5 from both sides and simplify.

\frac{x}{8}+5-5=42-5\\\\\frac{x}{8}=37\\\\

3. Multiply 8 on both sides and simplify.

8*\frac{x}{8}=37*8\\\\x=296

<u>We have found our number, it's 296!</u>

FrozenT [24]2 years ago
3 0

Answer:the number is 296

Step-by-step explanation:this is right cause I calculated in my head 296.

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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
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Answer:

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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

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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}}

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

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\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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