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4 years ago

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Books at a library sale are sold for $3.50 each. A function, y = 3.50x can be used to generate an input/output table for this sc

enario. Which statement is true about an accurate graph of this data? The data is discrete, so the points are connected with a line. The data is discrete, so the graph is a series of unconnected points. The data is continuous, so the graph is a series of unconnected points. The data is continuous, so it does not matter whether or not the points are in a line.

1 answer:

Thepotemich [5.8K]4 years ago

8 0

Answer:

the first one is the answer

Step-by-step explanation:

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Else purchased a prepaid phone card for $15. Long distance calls cost 6 cents a minute using this card. Elsa used her card only

Ede4ka [16]

Answer:

25 minutes

Step-by-step explanation:

First you subtract the original amount on the card from the remaining amount left.

So you can get the amount of money spent on that call.

$15 - $13.50 = $1.50

Then you divide that number by how much it cost per minute.

$1.50 ÷ 6 = 0.25

Then you get your answer!

She was on the phone for 25 minutes

7 0

3 years ago

Water is leaking from a jug at a constant rate. After leaking for 2 hours, the jug contains 48 fluid ounces of water. After leak

devlian [24]

Answer:

2 fluid ounce/hour

Step-by-step explanation:

2 hours to 5 hours ; 3 hours apart

leaks: 48 -42 = 6 fluid ounce

rate of leak = 6/3 = 2 fluid ounce/hour

8 0

4 years ago

Calculate how much each size per gram, and circle which gives the best value for money. 250g for $2.30, 400g for &3.40, 1Kg

Bezzdna [24]

To do this, we must set up ratios:
Option One:
$\frac{2.30}{250g} \\ \frac{0.0092}{1g} \\ 0.0092/g$
The first options costs only $0.0092/gram (a fraction of a penny)!
Option Two:
$\frac{3.40}{400g} \\ \frac{0.0085}{1g} \\ 0.0085/g$
The second option costs only $0.0085/gram (cheaper than option one)!
Option Three:
This requires a little more work. First, we have to convert the grams into kilograms. For every 1 kg, there is 1,000 g. Therefore, 1,000g costs $5.65. Next, we set up the ratio as usual:
$\frac{5.65}{1000g} \\ \frac{0.00565}{1g} \\ 0.00565/g$
The third option costs $0.00565/gram.

Therefore, option three is the cheapest!

Hope this helps!

4 0

4 years ago

You are facing South. You make an about face, then turn 90 degrees right. What direction are you facing?

Roman55 [17]

Answer:

west

Step-by-step explanation:

4 0

3 years ago

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

<u>These are the steps to find the Taylor series for the function</u> $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

<u>To find the first three non-zero terms you need to replace n=3 into the sum</u>

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$

<u>To find the interval on which the series converges you need to use the Ratio Test that says</u>

For the power series centered at x=a

$P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,$

suppose that $\lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.$ . Then

If $R=\infty,$ the the series converges for all x
If $0$ then the series converges for all $|x-a|$
If R=0, the the series converges only for x=a

So we need to evaluate this limit:

$\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |$

Simplifying we have:

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |$

Next we need to evaluate the limit

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}$

-(n+1)(2n+1) is negative when n -> ∞. Therefore $|-(n+1)(2n+1)}|=2n^{2}+3n+1$

You can use this infinity property $\lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty$ when a>0 and n is even. So

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty$

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is $( -\infty, \infty )$ .

6 0

3 years ago