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3 years ago

What is the distance between the points (–3, 4 and (–7, 4? a. 0 b. 3 c. 4 d. 10?

1 answer:

5 0

Try graphing it to help, but since it's not a slanted line (because the y-values are the same), you just do x1-x2:

-3 - -7 = 4

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There were 214 people at a concert. Main floor tickets cost $17 and balcony tickets cost $11 for a total of $2954. How may balco

earnstyle [38]

There is 100 $17 tickets no 114 $11 tickets

6 0

3 years ago

Write a complete two-column proof for the following information. Hint: Use the Angle Addition Theorem and the fact that a line i

kramer

Answer:

Step-by-step explanation:

Given: m∠1 = 62° and lines t and l intersect

Prove: m∠4 = 62°

Proof:

Statement Reason

m∠1 = 62° Given

m∠1 , m∠2 are supplementary t is a straight line hence linear pair.

m∠4 , m∠2 are supplementary r is a straight line hence linear pair.

Angle 2=180-62 = 118 Definition of supplementary angles

Angle 4 = 180-118 =62 -do-

Angle 1 = Angle 4 Equality property

Hence proved

6 0

2 years ago

Read 2 more answers

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

2 years ago

Find a formula for the general term an of the sequence, assuming that the pattern of the first few terms continues. (Assume that

Allushta [10]

Answer:

The general term for the sequence can be given by the following formula:

$a_n=2\,n+9$

Step-by-step explanation:

If the sequence you typed starts with first term 11 and continues with terms 13, 15, 17, 19, We understand that the sequence is formed by adding 2 units to the previous term. So we are in the case of an arithmetic sequence with constant difference (d) = 2, and with first term 11.

Therefore, the nth term of this arithmetic sequence can be expressed by using the general form for an arithmetic sequence as:

$a_n=a_1\,+\,(n-1)\,d\\a_n=11\,+\,(n-1)\,2\\a_n=11+2\,n-2\\a_n=2\,n+9$

5 0

2 years ago

In Buffalo, New York the temperature was -14 degrees in the morning. if the temperature dropped 7 degrees. what is the degrees n

Veseljchak [2.6K]

Answer:

-21

Step-by-step explanation:

-14 - 7 = -21

4 0

2 years ago