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3 years ago

Two numbers total 89 and have a difference 0f 13. find the two numbers

Mathematics

1 answer:

attashe74 [19]3 years ago

5 0

X + Y = 89
X - Y = 13

So, X = 13 +Y

(13 +Y) + Y = 89
2Y = 89 - 13
Y = 38
X= 51 (13+38)

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Plz answer for number 12-18 it homework

solniwko [45]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

The 12th term and the sim of 1+3+5+7+.....

zzz [600]

To find this, first find the factor or rate of which the numbers are moving. To do so do as follows.

subtract 1 from 3

3-1=2

So each number is having 2 added to it.

Now add two to 7 and the numbers afterwards till you get the 12th term

7+2=9

1+3+5+7+9

9+2=11

1+3+5+7+9+11

11+2=13

1+3+5+7+9+11+13

13+2=15

1+3+5+7+9+11+13+15

15+2=17

1+3+5+7+9+11+13+15+17

17+2=19

1+3+5+7+9+11+13+15+17+19

19+2=21

1+3+5+7+9+11+13+15+17+19+21

21+2=23

1+3+5+7+9+11+13+15+17+19+21+23

So 23 is the 12th term

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4 years ago

A circle cuts the x-axis at (4,0) and (14,0), and cuts the y-axis at (0,6) and (0,8). Find it's equation

gayaneshka [121]

It is not a circle but you can find the equation of the ellipse.
To do this we need to work out the major and minor radii and the centre
The centre is at (9, 7)
The major (y) radius is 1 and the minor (x) radius is 5
Therefore the equation is (x-9)/5 + (y-7)/1 = 1

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3 years ago

Find the oth term of the geometric sequence 7, 14, 28, ...

yaroslaw [1]

Answer:

The nth term of the geometric sequence 7, 14, 28, ... is:

$a_n=7\cdot \:2^{n-1}$

Step-by-step explanation:

Given the geometric sequence

7, 14, 28, ...

We know that a geometric sequence has a constant ratio 'r' and is defined by

$a_n=a_1\cdot r^{n-1}$

where a₁ is the first term and r is the common ratio

Computing the ratios of all the adjacent terms

$\frac{14}{7}=2,\:\quad \frac{28}{14}=2$

The ratio of all the adjacent terms is the same and equal to

$r=2$

now substituting r = 2 and a₁ = 7 in the nth term

$a_n=a_1\cdot r^{n-1}$

$a_n=7\cdot \:2^{n-1}$

Therefore, the nth term of the geometric sequence 7, 14, 28, ... is:

$a_n=7\cdot \:2^{n-1}$

6 0

3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago