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aksik [14]
3 years ago
6

First right answer gets brainliest

Mathematics
2 answers:
Finger [1]3 years ago
5 0
Problem:
find line perpendicular to 4x+7y+3=0 passing through (-2,1).

Solution:
We will first find the general form of the perpendicular line.
The perpendicular line has the form
7x-4y+k=0 .........................(1)
by switching the coefficients of x and y, and switching the sign of one of the two coefficients.  This way, the slope of (1) multiplied by that of the original equation is -1, a condition that the two lines are perpendicular. The value of k is to be determined from the given point (-2,1).

To find k, we substitute x=-2, y=1 into equation (1) and solve for k.
7(-2)-4(1)+k=0
=>
k=14+4=18
Therefore the required line is
7x-4y+18=0

blsea [12.9K]3 years ago
5 0
If you know the slope of a line and a point the line passes through, you can find the equation of the line using the formula:

y - y_1 = m(x - x_1)

You are given a point the line passes through, (-2, 1).
Now you need to slope of your line.
You are given the equation of a line that your line is perpendicular to.
The slopes of perpendicular lines are negative reciprocals.
If you find the slope of the given line, you can find the slope of the perpendicular line.

Given line:

4x + 7y + 3 = 0

Solve for y:

7y = -4x - 3

y = (-4/7)x - 3/7

The slope of the given line is -4/7.
The slope of the perpendicular is 7/4.

Now we can find the equation of the line using the given point and the slope we just found.

y - y_1 = m(x - x_1)

y - 1 = (7/4)(x - (-2))

y - 1 = (7/4)(x + 2)

4y - 4 = 7x + 14

7x - 4y + 18 = 0
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Answer: 5 is your answer hope this helped

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Step-by-step explanation:

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What is 62.4 as a fraction and simplified?
kumpel [21]

Answer:

62 2/5

Step-by-step explanation:

62 is just the whole number and then you have 4 tenths, which is 4/10 which can be simplified to 2/5, so it's 62 2/5

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Write 99.124 in expanded form
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The answer that you are looking for is
90+9+0.1+0.02+0.004
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a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter
andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|, converges if 0< b_{n+1} \leq b_n for all n, and \lim_{n \to \infty} b_n = 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

<em>Part a</em>

An Alternating series is an infinite series given on these three possible general forms given by:

\sum_{n=0}^{\infty} (-1)^{n} b_n

\sum_{n=0}^{\infty} (-1)^{n+1} b_n

\sum_{n=0}^{\infty} (-1)^{n-1} b_n

For all a_n >0, \forall n

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

<em>Part b</em>

An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|  converges if 0< b_{n+1} \leq b_n for all n and \lim_{n \to \infty} b_n =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

<em>Theorem (Alternating series test)</em>

If a sequence of positive terms {bn} is monotonically decreasing and

<em>\lim_{n \to \infty} b_n = 0<em>, then the alternating series \sum (-1)^{n-1} b_n converges if:</em></em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

then <em>\sum_{n=1}^{\infty}(-1)^{n-1} b_n  converges</em>

<em>Proof</em>

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

|s_n -s| < \epsilon for all n, and that implies that the series converges to a value, s.

And this complete the proof.

<em>Part c</em>

An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series

By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:

Rn = s - sn

Where s_n represent the partial sum for the series and s the total for the sum.

Is important to notice that the size of the error is at most b_{n+1} by the following theorem:

<em>Theorem (Alternating series sum estimation)</em>

<em>If  \sum (-1)^{n-1} b_n  is the sum of an alternating series that satisfies</em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

Then then \mid s - s_n \mid \leq b_{n+1}

<em>Proof</em>

In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any  two consecutive partial sums sn and sn+1.

\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}

And this complete the proof.

5 0
4 years ago
Eloise is investing in a retirement account. She plans on adding an additional $50 at the end of every year and the expected mon
Lisa [10]
The accumulated (future) value is given by the formula
F=P(1+i)^n
where 
P=amount of deposit (made at the beginning of the first period)
i=monthly interest, APR/12 = 3%/12 =0.0025
n=number of periods (month)

For example, the future value for the 6th month is
F(6)=1000(1.0025^6)=1015.09  (to the nearest cent)

Here is a schedule of the values,
i=month
F(i) = value at the end of month i.

  i   F(i)
 0  1000.0 
 1  1002.5 
 2  1005.01
 3  1007.52
 4  1010.04
 5  1012.56
 6  1015.09
 7  1017.63
 8  1020.18
 9  1022.73
10 1025.28
11 1027.85
12 1030.42 + $50 deposit = 1050.42
All values are rounded to the nearest cent.
6 0
3 years ago
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