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Yanka [14]
2 years ago
13

11- pls help!!!! math - -

Mathematics
1 answer:
Andrej [43]2 years ago
7 0

Answer:

B

Step-by-step explanation:

-1*2+3*5 =13

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What is the smallest possible degree for the polynomial function above? Explain your reasoning.
Blababa [14]

Answer:

(Yes, "5" is a polynomial, one term is allowed, and it can be just a constant!) 3xy-2 is not, because the exponent is "-2" (exponents can only be 0,1,2,...)

Step-by-step explanation:

8 0
3 years ago
From a standard deck of 52 cards, what is the probability of picking a Heart at random from the deck? A) 1 13 B) 1 26 C) 1 4 D)
monitta

Answer:

C 1/4

Step-by-step explanation:

13/52 = 1/4

6 0
4 years ago
Hence, find the smallest possible value of z such that 160 x z is a perfect square​
vodka [1.7K]

Answer:

10.

Step-by-step explanation:

I am assuming that z is a whole number.

160z = x^2

z = x^2/ 160.

160 = 2*2*2*2*2*5

To make this a perfect square we multiply by 2 * 5

- this gives us 1600.

z = 1600/160 = 10.

5 0
3 years ago
A swimming pool is in the shape of a right triangle. One leg has a length of 10 feet and one leg has a length of 15 feet. Estima
sasho [114]

Answer:

5√13 = c  OR 18. 0

Step-by-step explanation:

Using the Pyhtagorean Theorem, we will solve for x (in this case our hypothenuse)

a2 + b2 = c2

10(square) + 15(square)= c2  .....10*10 = 100  and   15* 15= 225

100 + 225 = c2

325= c2

√325 = c

5√13 = c  OR 18. 0

5 0
3 years ago
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
FromTheMoon [43]

Answer:

The Taylor series is \ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

The radius of convergence is R=3.

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at a=3 its expression is given in powers of (x-3). With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of \ln(1+x).

Then,

\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).

Now, in order to make a more compact notation write \frac{x-3}{3}=y. Thus, the above expression becomes

\ln(x) = \ln 3 + \ln(1+y).

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.

Now, substitute \frac{x-3}{3}=y in the previous equality. Thus,

\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},

Where a_n stands for the coefficients of the Taylor series and R for the radius of convergence.

In this case the coefficients of the Taylor series are

a_n = \frac{(-1)^{n+1}}{ n3^n}

and in consequence |a_n| = \frac{1}{3^nn}. Then,

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}

Applying the properties of roots

\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.

Hence,

R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}

Recall that

\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.

So, as R^{-1}=\frac{1}{3} we get that R=3.

8 0
4 years ago
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