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Monica [59]
3 years ago
9

Please help . Plot the points & determine the slope

Mathematics
1 answer:
Marysya12 [62]3 years ago
6 0
The slope is 3/1 you plot the points then you go up 3 and over 1
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What is the first step in solving the equation 9m - 2 = 16?
ElenaW [278]

Answer: Add 2 to each side

Step-by-step explanation:

9m-2=16

+2 +2

9m=18

8 0
3 years ago
Read 2 more answers
If f(x) = 3 ^ x - 7 and g(x) = x ^ 2 + 11 , the value of f(2) - g(7) is:
Maurinko [17]

Answer:

The value of f(2) - g(7)=-58

Step-by-step explanation:

We have

f(x) = 3 ^ x - 7\\\\f(2) = 3 ^ 2 - 7\\\\=9-7\\\\=2\\\\g(x) = x ^ 2 + 11\\\\g(7) = 7 ^ 2 + 11\\\\=49+11\\\\=60\\\\f(2) - g(7)=2-60\\\\=-58

4 0
3 years ago
A lines y-intercept is -4, and it's slope is 2. what is its equation in slope-intercept form?
borishaifa [10]

Answer:

2x-4

Step-by-step explanation:

2x-4

mx+b format

3 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
There are 48 calories of energy in 40g of rice. How many calories are in 160 g of rice?
Rashid [163]
160 grams of rice is four times larger than the given 40 g. 
<span>So the number of Calories must be 4 times larger too. </span>

<span>48 x 4 = 192 Calories in 160 g of rice</span>
5 0
3 years ago
Read 2 more answers
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