What operation do you perform first when you simplify 175 subtracting (8+45÷3)×7

Mathematics

1 answer:

Zepler [3.9K]3 years ago

4 0

175 - (8+45÷3)×7

Using BIDMAS

= 175 - (8+15)×7
= 175 - (23×7)
= 175 - 161
= 14

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Out or 1,000,000,000.00, how many 10,000.00s are there?

yawa3891 [41]

Answer:

100,000.00

Step-by-step explanation:

Divide 1,000,000,000.00 by 10,000.00

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SVETLANKA909090 [29]

1) given that those two lines are parallel, we can deduce that 9/45=x/75.
9/45 simplifies to 1/5 so 1/5 = x/75. Cross multiply to get this: 5x = 75.
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2) Again, we can set up a proportion to solve for x because we know the lines are parallel. 24/36=x/12. 24/36 simplifies to 2/3 so 2/3=x/12. Cross multiply to get this: 3x = 24. Divide both sides by 3 to get x.
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4 years ago

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

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

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Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .