All categories

2 years ago

How are dilations and functions related?

2 answers:

5 0

If we dialate a function we are just changing what the graph looks like by a stretch or a compression, the function really doesn't change. If a function before the dilation, then it is a function after the dilation. The opposite is also true.

loris [4]2 years ago

4 0

Answer:

A function is said to be dilated when it has been stretched away from an axis or compressed toward an axis. This dilation can be a vertical or a horizontal stretch.

When the given function is multiplied by any number, it will automatically dilate but at the same time, will also retain its basic shape.

You might be interested in

What is 100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Allisa [31]

Answer:

Step-by-step explanation:

;)

3 0

2 years ago

What is 0.05 divided by 4.8 ?

postnew [5]

I Got 0.01041666666 As An Answer <span />

5 0

3 years ago

Read 2 more answers

Please help me find what y equals i need help, thank you so much, god bless you.

il63 [147K]

Answer:

Step-by-step explanation:

2(4)=8

8+7=15

hope this helped, have a blessed day!

4 0

2 years ago

Read 2 more answers

25 is 32 percent of what number?

Reptile [31]

The answer is: " 0.78125 " .
_____________________________________________
32% = 32/100 = 32÷100 = 0.32 ;

(0.32)n = 25 ; solve for "n" ;

32 n = 25 ;

32 n / 32 = 25/ 32 ;

n = 25/32 = 0.78125
___________________________________________________

not applicable/ not an answer/ non-answer to: "non-question and/or incomplete question content;
absurd/confusing/incomplete/improper answer/ please review! THANK YOU! Kind regards.

8 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cbegin%7Bequation%7D%5Ctext%20%7B%20Question%3A%20If%20%7D%20%5Cint_%7B%5Cfrac%7B-1%7D%7B%5C

Elodia [21]

We want to evaluate

$\displaystyle \int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \sqrt{\left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2} \, dx$

First we note that the integrand is even (replacing x with -x doesn't fundamentally alter the function being integrated), so this is equal to

$\displaystyle 2 \int_0^{\frac1{\sqrt2}} \sqrt{\left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2} \, dx$

The radicand reduces significantly to

$\displaystyle \left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2 = \frac{16x^2}{(1-x^2)^2}$

so that taking the square root, we simplify the integral to

$\displaystyle 8 \int_0^{\frac1{\sqrt2}} \frac x{1-x^2} \, dx$

which is trivially computed with a substitution of $y = 1 - x^2$ and $dy=-2x\,dx$ :

$\displaystyle -4 \int_1^{\frac12} \frac{dy}y = 4 \int_{\frac12}^1 \frac{dy}y = -4 \ln\left(\frac12\right) = \ln(2^4) = \boxed{\ln(16)}$

7 0

1 year ago