3 years ago

So basically I need help

Mathematics

1 answer:

vovangra [49]3 years ago

7 0

Um- thats hard OH YOU WANT 20 CHARACTERS ILL GIVE YOU 20 CHARACTERS STOP SAYING THAY

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Simplify -12x^4/x^4+8x^5

alexgriva [62]

Answer:

OPTION A: $\frac{12}{1 + 8x}$ , where $x \ne - \frac{1}{8}$ .

Step-by-step explanation:

Given: $\frac{- 12x^4}{x^4 + 8x^5}$

Taking $x^4$ common outside in the denominator, we get:

$= \frac{-12x^4}{(x^4)(1 + 8x)}$

$x^4$ will get cancelled on the numerator and denominator, we get:

$= \frac{-12}{1 + 8x}$

we know that the denominator can not be zero.

That means, 1 + 8x $\ne$ 0.

$\implies 8x \ne -1$

$\implies x \ne \frac{-1}{8}$

So, the answer is: $\frac{-12}{1 + 8x}$ , where $x \ne \frac{-1}{8}$ .

6 0

4 years ago

By rounding to 1 significant figure, estimate <br><br> 306.9 x 6.4

coldgirl [10]

Answer:2000

Step-by-step explanation:

mutiply

306.9 * 6.4

=1964.16

to one significant figure

=2000

8 0

3 years ago

Help me answer a-d please it do tomorrow

Papessa [141]

A) $320
b) 5 weeks
c) $80

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3 years ago

For the function defined by f(x)=12/x^2, what is the constant proportionality

viva [34]

Answer:

It's C, 12

Step-by-step explanation:

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4 years ago

What’s the answer of g

Lunna [17]

Answer:

$\huge\boxed{g(a-1)=\dfrac{2-a}{a}=\dfrac{2}{a}-1}$

Step-by-step explanation:

$g(t)=\dfrac{1-t}{1+t}\\\\g(a-1)-\text{substitute}\ t=a-1\ \text{to}\ g(t):\\\\g(a-1)=\dfrac{1-(a-1)}{1+(a-1)}=\dfrac{1-a-(-1)}{1+a+(-1)}=\dfrac{1-a+1}{1+a-1}=\dfrac{(1+1)-a}{(1-1)+a}=\dfrac{2-a}{a}$

6 0

3 years ago