3 years ago

If 12 to 99 power - 12 to 97th power

Mathematics

2 answers:

mariarad [96]3 years ago

8 0

Answer:

Step-by-step explanation:

12⁹⁹ - 12⁹⁷ = 12⁹⁷⁺² - 12⁹⁷

= 12⁹⁷*12² - 12⁹⁷

= 12⁹⁷* [ 12² - 1]

= 12⁹⁷ * [ 144 - 1]

= 12⁹⁷ * 143

Sindrei [870]3 years ago

3 0

Answer:

The question is a little confusing but I am going to assume that you want 12 to the 97th power and that is it.

Step-by-step explanation:

. 4.792707e+104 is the answer. I just put it in the calculator. But what you do us 12x12x12x12 and so on 97 times. Hope this helps!! Have a good day:-)

You might be interested in

Suppose you have just inherited a sum of money. Let’s say you choose to invest the money. How much did you inherit? Choose an am

VladimirAG [237]

Answer:

What is your interest rate?

Step-by-step explanation:

6 0

2 years ago

Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for

USPshnik [31]

$x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)\\\\-x^2-3x+4=-(x^2+3x-4)=-(x^2+4x-x-4)\\\\=-[x(x+4)-1(x+4)]=-(x+4)(x-1)\\\\f(x)=\dfrac{x^2+7x+12}{-x^2-3x+4}=\dfrac{(x+4)(x+3)}{-(x+4)(x-1)}\\\\\text{Vertical asymptotes:}\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\\boxed{x=-4\ and\ x=1}\\\\\text{Horizontal asymptotes:}$

$\lim\limits_{x\to\pm\infty}\dfrac{x^2+7x+12}{-x^2-3x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\dfrac{7}{x}+\dfrac{12}{x^2}\right)}{x^2\left(-1-\dfrac{3}{x}+\dfrac{4}{x^2}\right)}=\dfrac{1}{-1}=-1\\\\\boxed{y=-1}$