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2 years ago

Rational Numbers-

Mathematics

1 answer:

ioda2 years ago

5 0

Answer:

-9 11/15

2. 11/24

34/45

4. -5.05

5. -0.622

Step-by-step explanation:

1. -2 4/5 - 6 2/15

convert them to a common denominator and an improper fraction:

-2 4/5= 18/5=- 54/15 -6 2/15= 92/ 15

*15 is the other denominator, so you keep the numbers in the second number the same*

subtract: -54/15- 92/15

since you are subtracting a negative number it goes farther away from zero

-146/15 or -9 11/15

2. -5/12 + 7/8

convert them to a common denominator

-5/12=- 20/48 7/8= 35/40

add: -20/48+35/40

11/24

since you are adding a negative number it goes closer to zero or past zero:

3. 3 3/5 x 2/9

convert 3 3/5 to an improper fraction: 18/5

mulitply the two denominators and the two numerators

17/5 x 2/9= 34/45

This is also the simplest form.

4. Convert to a decimal. -5 1/20

keep the -5 beacuse that is a whole number

divide 1/20 to turn it to a decimal to get 0.05

5. Which number is the largest? -0.65, -0.622, -5/8

to find this you would have to convert all numbers to either a decimal or a fraction.

5/8 is 0.625

the one closest to zero is the largest which is -0.622

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3 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

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so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

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3 years ago

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Answer:

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