All categories

3 years ago

What is 0.004 divided by 0.05

Mathematics

1 answer:

Korvikt [17]3 years ago

4 0

When u divide 0.004 by 0.05 you will get 0.08 as the quotient.

<span />

You might be interested in

What's the simplist way to simplify 13/78

Andrews [41]

13/78 simplified is 1/6

4 0

4 years ago

Read 2 more answers

Consider an urn with 7 black balls, 3 yellow balls, and 4 orange balls. If 3 balls are chosen randomly with replacement, what is

KATRIN_1 [288]

Answer:

1.79% chance or $\frac{6}{343}$

Step-by-step explanation:

7+3+4=14

chance of yellow $\frac{3}{14}$

chance of orange $\frac{4}{14}$

$\frac{3}{14} * \frac{4}{14} *\frac{4}{14} = \frac{6}{343}$

5 0

2 years ago

Rewrite the equations using a negative exponent!! ill give brainliest, any random or incomplete answers will be reported!!

vesna_86 [32]

Answer:

11. $x^{-2}$

12. $m^{-3}$

13. $(x+2y)^{-4}$

14. $(x+3y)^{-2}$

15. $(x+y)^{-3}$

16. $(2x+3y)^{-6}$

17. $(-3m-9)^{-3}$

18. $x^{-4}$

Step-by-step explanation:

In all of them you just subtract the bottom exponent from the top exponent.

5 0

3 years ago

Read 2 more answers

I need helpppp with 1-9

BigorU [14]

Hi!! What do you need help with? Solving the problems? Or understanding them? Also 1-9 is a bit blurry, so if you could retake it and repost it, I could solve for them. :). But if you don’t want to, don’t worry.

3 0

3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago