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3 years ago

Using the digits 5, 6, 7, 8, 9 and repetition is allowed how many options are there to

Mathematics

1 answer:

just olya [345]3 years ago

8 0

Answer:

250 options to create an even number

375 options to create an odd number

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Trip has 15 coins worth 95 cents.Four of the coins are each worth twice as much as the rest. Construct a math argument to justif

const2013 [10]

(11*5)+(4*10)=95
11+4=15
Both are equal to the constraints of the word problem and there for justified.

6 0

3 years ago

A jar contains n nickels and d dimes. There are 24 coins in the jar, and the total value of the coins is $1.90. How many nickels

oee [108]

Answer:14 dimes 10 nickels

Step-by-step explanation:

24 coins total

14 X.10 = 1.40

10 x .05 = .50

1.40 + .50 = 1.90

3 0

3 years ago

Work out the percentage change to 2 decimal places when a price of £198 is increased to £209.99.

bija089 [108]

Answer:

The percentage change = 6.05%

Step-by-step explanation:

New value = £209.99

Old value = £198

Percentage change = [New Value - Old Value] / [Old Value] × 100

= [209.99 - 198] / [198] × 100

= [11.99] / [ 198] × 100

= 0.06 × 100

= 6.05%

Therefore, the percentage change = 6.05%

5 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

A business could not collect $5,000 that it was owed. The total owed to the business was $100,000. What fraction of the total wa

son4ous [18]

Total owed to the business = $100,000

amount that could not be collected = $5000

Fraction of total not collected

$\text{fraction not collected=}\frac{5000}{100000}=\frac{5}{100}=\frac{1}{20}$

3 0

1 year ago