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3 years ago

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James researched gas prices for the last 3 years (as shown in the table below). Based on your line of best fit, what will be the

price of gas 50 years from now? (Hint: year 53)
A. $12.75

B. $14.23

C. $15.42

D. $17.99

1 answer:

kykrilka [37]3 years ago

6 0

Most likely C.$15.42

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Helene is finding the sum (9 10i) (–8 11i). She rewrites the sum as (–8 11)i (9 10)i. Which statement explains the property of a

Andreyy89

She made the mistake of grouping unlike terms and factorizing.

Given that

Helene is finding the sum (9 + 10i) + (–8 + 11i).

She rewrites the sum as (–8 + 11)i + (9 + 10)i.

We have to determine

Which statement explains the property of addition that she made an error in using?

According to the question

The mistake she did is in the second term distributing.

(9+10i) is not equal to (9+10)i

Similarly (-8+11i) is not equal to (-8+11)i.

The correct method she should have done is given below;

Grouping real terms together and imaginary terms together and finding the sum is,

$\rm = (9 + 10i) + (-8 + 11i)\\\\= 9+10i-8+11i\\\\= (9-8) +(10i+11i)\\\\=1+21i$

Hence, she made the mistake of grouping unlike terms and factorizing.

To know more about Complex Number click the link given below.

brainly.com/question/10078818

6 0

2 years ago

I need answers to all of these. Please help!! I suck at math..

RSB [31]

1)  (-4) + (-9) = -13

2) 5 + (-12) = -7

3) -6 + 5 = -1

4) -7 + 7 = 0

<span>5) -7 + 2 ___ 3 + (-9)

- 5 ____ - 6

</span>-7 + 2 >  3 + (-9)

6)  <span>4 + (-8) ___ 6 + (-8) + (-6)

- 4 ____ - 8

</span><span>4 + (-8) >  6 + (-8) + (-6)

7)  </span><span>$66

8)  </span><span>-$12
</span>

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3 years ago

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

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

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

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3 years ago

Which statement is not necessarily true?<br> Given: DE is the ⊥ bisector of JL

Ket [755]

"DJ= DL" is the one statement among the following choices given in the question that is not necessarily true. The correct option among all the options that are given in the question is the fourth option. I hope that this is the answer that you were looking for and it has actually come to your desired help.

<u>THIS ANSWER IS WRONG EVERYBODY</u>

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3 years ago

12 days of math challenge

erastova [34]

Answer:

Reindeer and snowflake are 4

tree is 8

snowman is 1

Step-by-step explanation:

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2 years ago