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2 years ago

14

Point F' corresponds to point D. point E. point F. opoint E.

2 answers:

Scilla [17]2 years ago

7 0

After transformation the Point F corresponds to :

Point F'

because they both form right angles.

Arada [10]2 years ago

6 0

refer to the attachment

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A stadium has 49000 seats. Seats sell for $25 in section A, $20 in section B, and $15 in section C. The number of seats in secti

eimsori [14]

Answer:

A stadium has 49000 seats.

Seats sell for $25 in Section A, z

$20 in Section B,-------------------x seats

$15 in Section C. ------------------y seats

(x+y)=z

25(x+y)+20x+15y=1052000

25x+25y+20x+15y=1052000

45x+40y=1052000

/5

9x+8y=210400------------------1

2x+2y=49000

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Step-by-step explanation:

4 0

3 years ago

Zane calculated the missing side length of one of these triangles using the Pythagorean Theorem. Which triangle is it?

topjm [15]

The pythagorean theorem states that for any RIGHT-ANGLED TRIANGLE, the relationship between the side lengths are A²+B²=C², where C is the hypotenuse :P

Which means that it can only be applied on right angled triangles~
So your answer would be B, a right-angled triangle.

Ask me if you are unsure about anything <3

6 0

2 years ago

Read 2 more answers

13x - 3<br> 4.<br> 6x +<br> 1<br> 68

Rasek [7]

Answer:

= -21.6x + 168

Step-by-step explanation:

Combine Like Terms

13x -34.6x + 168

13x + - 34.6x + 168

(13x+ -34.6x) + 168

-21.6x + 168

3 0

2 years ago

Why might algebra tiles not be a good tool to use to factor x2 + 18x + 80?

kvasek [131]

<span>
factor a trinomial to (10+x) (8+x)
hope it helps</span>

7 0

2 years ago

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