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3 years ago

Solve a system of equations using substitution: word problems

1 answer:

4 0

Restaurant 1 : y = 7k + 39
restaurant 2 : y = 2k + 44

7k + 39 = 2k + 44
7k - 2k = 44 - 39
5k = 5
k = 5/5
k = 1

y = 2k + 44
y = 2(1) + 44
y = 2 + 44
y = 46

Dinner would cost $ 46 at either restaurant if Lina's lobster weighed 1 kilogram.

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

musickatia [10]

Answer:

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

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

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).

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

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

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3 years ago

What is the value of s in the equation below? S + 1 1/2 = 10 1/2

LenaWriter [7]

Answer:

S=9

Step-by-step explanation:

10 $\frac{1}{2}$ - 1 $\frac{1}{2}$ = 9

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emmasim [6.3K]

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3 years ago

Will give brain less

irina1246 [14]

Answer:

(0, 0)

Step-by-step explanation:

The y-intercept refers to the point where the graph of a function intersects or crosses the y-axis. The value of x is usually zero at this particular point since the y-axis has the equation;

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