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emmasim [6.3K]
3 years ago
12

Please help me on that question for my Algebra. I don't understand it.

Mathematics
2 answers:
tigry1 [53]3 years ago
5 0
13 quarters = 13 *0.25 = 3.25

3.60 - 3.25 = 0.35

0.35 = 0.25 +0.10

he has 1 dime and 14 quarters

Answer is B
fgiga [73]3 years ago
3 0
14 quarters x .25 = 3.50
1 dime is .10
3.50 + .10 = 3.60
B. 1 Is your answer.
Your pfp looks amazing btw.

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Roberto will save 1/6 of his allowance each day. If he gets $2.00 a day, about how much money will he save each day? Round your
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Amount saved= fraction saved x amount of money received
A= 1/6 x 2
A=1/3
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33 cents or $.33

Hope this helps :)
6 0
3 years ago
is (3,-3) a solution to the system 5x+4y=3, 2x-5y=19 need to be taught step by step please and thank you.
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Basically all you’re doing is plugging in the inputs to the equations and see if they’re correct.

5(3)+4(-3)=3 this proves to be true
2(3)-5(-3)=21 not 19
8 0
3 years ago
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According to the rational root theorem, the
nydimaria [60]

Answer:

Step-by-step explanation:

Sorry this is a tough one and I don’t want to give you the wrong answers sorry again

6 0
3 years ago
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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
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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3 years ago
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The temperature on Tuesday afternoon is below 38 degrees Fahrenheit. As it states in the question the temperature on Tuesday morning was 27 degrees Fahrenheit and rose 10 more degrees at noon. Which makes is 37 degrees Fahrenheit at noon on Tuesday. You get this by taking 27 degrees and adding 10 degrees.  
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