Y-1/x+3=-3/4 how do you solve for y

1 answer:

3 0

$\frac{ y-1}{x+3} =- \frac{3}{4}\\ \\x+3\neq 0\\ \\x\neq -3 \\ \\D=R\setminus \left \{ -3 \right \}$

$4(y-1)=3(x+3)\\ \\4y-4=3x+9\\ \\4y=3x+9+4\\ \\4y=3x+13\ \/:4\\ \\y=\frac{3}{4}x +\frac{13}{4}$

You might be interested in

A person earns $187.42 before deductions for 3 days' work. What is the rate of pay per hour assuming 8 hours of work per day?

gavmur [86]

8 hours = 1 day

24 hours = 3 days

187.42/24

Round it up and it will give you $7.80/hour

8 0

3 years ago

How do the values in Pascal’s triangle connect to the coefficients?

damaskus [11]

Explanation:

Each row in Pascal's triangle is a listing of the values of nCk = n!/(k!(n-k)!) for some fixed n and k in the range 0 to n. nCk is <em>the number of combinations of n things taken k at a time</em>.

If you consider what happens when you multiply out the product (a +b)^n, you can see where the coefficients nCk come from. For example, consider the cube ...

(a +b)^3 = (a +b)(a +b)(a +b)

The highest-degree "a" term will be a^3, the result of multiplying together the first terms of each of the binomials.

The term a^b will have a coefficient that reflects the sum of all the ways you can get a^b by multiplying different combinations of the terms. Here they are ...

(a +_)(a +_)(_ +b) = a·a·b = a^2b
(a +_)(_ +b)(a +_) = a·b·a = a^2b
(_ +b)(a +_)(a +_) = b·a·a = a^2b

Adding these three products together gives 3a^2b, the second term of the expansion.

For this cubic, the third term of the expansion is the sum of the ways you can get ab^2. It is essentially what is shown above, but with "a" and "b" swapped. Hence, there are 3 combinations, and the total is 3ab^2.

Of course, there is only one way to get b^3.

So the expansion of the cube (a+b)^3 is ...

(a +b)^3 = a^3 + 3a^2b +3ab^2 +b^3 . . . . . with coefficients 1, 3, 3, 1 matching the 4th row of Pascal's triangle.

In short, the values in Pascal's triangle are the values of the number of combinations of n things taken k at a time. The coefficients of a binomial expansion are also the number of combinations of n things taken k at a time. Each term of the expansion of (a+b)^n is of the form (nCk)·a^(n-k)·b^k for k =0 to n.