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3 years ago

Rounded to the nearest million \$58,788,188

Mathematics

1 answer:

almond37 [142]3 years ago

6 0

59,000,000 is your answer now if it was nearest 10 millions it would be 60,000,000

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Why do we need to know about unit rates

allochka39001 [22]

Answer:

because they allow us to make direct comparisons

Step-by-step explanation:

8 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

Evaluate expression |3|+-19|

tankabanditka [31]

Answer:

If it suppose to be |3+-19| then the answer is 16

Step-by-step explanation:

4 0

3 years ago

Compare and Contrast:

algol [13]

Equation 1 :
| 5x - 6 | = -41.....no solution because an absolute vlue cannot equal a negative number

equation 2 :
| 7x + 13 | = 27...2 solutions

C. equation 2 has more solutions then equation 1.

6 0

3 years ago

Read 2 more answers

Given 2x+4y=12 what is the x and y int

Setler [38]

Both are 2! 2x2=4 4x2=8 so 8+4=12

3 0

3 years ago