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2 years ago

What is y=x -4; (5,1)

Mathematics

1 answer:

Ilya [14]2 years ago

5 0

1=5-4 because you have to substitute x and y (x,y) or (5,1)

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Draw a polygon with the given conditions in a coordinate plane of rectangle with a perimeter of 20 units

Anarel [89]

Question says to draw a polygon with the given conditions in a coordinate plane. It is talking about Rectangle with perimeter of 20 units.

So we can use formula of perimeter of rectangel to find it's possible dimensions.

We knwo that perimeter of the rectangle is given by 2(L+W)

where L is lenght and W is width.

Given perimeter is 20 so we get:

2(L+W)=20

L+W=10

L=10-W

So basically we can use any number for W which can be from 0 to 10 as side length can't be negative.

Let W=4

Then L=10-W=10-4=6

Hence we just have to draw a rectangle having length 6 and width 4.

So the final answer will be picture in the attached graph.

3 0

2 years ago

How many pairs of points are reflections across the x-axis? A)1 B)2 C)3 D)4

asambeis [7]

Answer:

C. 3

Step-by-step explanation:

C&D, A&F, K&J

6 0

3 years ago

The sum of three consecutive even integers is -78, what are the integers ?

Digiron [165]

First, you need to analize and understand the problem, then you must choose and strategy. There is a wide variety of strategies to solve a mathematical problem, in this case, you can use the following, which is based on the information given above:
1. You have that the sum of three consecutive even integers is $-78$ , therefore, you can given the variable $x
$ to the first integer, the second even integer is $x+2$ and the third one is $x+4
$ .
2. Calculate x:
 $x+(x+2)+(x+4)=-78$ 
 $3x+6=-78$
 $x=-28
$ 
 $x+2=-28+2=-26
$ 
 $x+4=-28+4=-24
$ 
Therefore, the answer is: $-28,-26,-24
$

5 0

2 years ago

Calculate the perimeter of the figure shown above:

larisa86 [58]

The answer is B, 50ft

8 0

3 years ago

Read 2 more answers

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago