$\dfrac{-14x^3}{x^3-5x^4}\qquad\text{where}\ x^3-5x^4\neq0\\\\x^3-5x^4\neq0\qquad\text{distributive}\\\\x^3(1-5x)\neq0\iff x^3\neq0\ \wedge\ 1-5x\neq0\\\\x\neq0\ \wedge\ x\neq\dfrac{1}{5}\\\\\dfrac{-14x^3}{x^3(1-5x)}\qquad\text{cancel}\ x^3\\\\=\dfrac{-14}{1-5x}\qquad\text{where}\ x\neq\dfrac{1}{5}$

6 0

3 years ago

Take 9 from 6, 10 from 9, 50 from 40 and leave 6. What is the answer?

tekilochka [14]

Take 9 from 6 = -3
Take 10 from 9 = -1
Take 50 from 40 and leave 6 = -4

7 0

3 years ago

Given log32≈0.631 and log37≈1.771, what is log314

Solnce55 [7]

Answer:

the log to the base 3 of 14 is 2.402

Step-by-step explanation:

You must find a way to indicate that 3 is the base; you cannot run this '3' together with 2, 7 or 14.

Example:

log to the base 3 of 2 = 0.631

log to the base 3 of 7 = 1.771

Note that 2 times 7 is 14. Thus, to obtain the log to the base 3 of 14, we must ADD the two logs shown above:

0.631

+1.771

----------

2.402

Thus, the log to the base 3 of 14 is 2.402.

Check: Does 3^2.402 = 14? YES

8 0

3 years ago