What is the solution to the system of equations below? Y=-4/5x-6 & y=-30

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

6 0

its A) 45,-30

i got this answer by using a graphing calculator.

Ainat [17]3 years ago

5 0

So when you are trying to find the solution to the system of equations make sure that you have it where y= -4/5x-6 and y= -30 are written like this:

-30=-4/5x-6

And then solve for x:

Add 6 on both sides.

-24=-4/5x

Then multiply by the reciprocal of -4/5.

-5/4*-24=-5/4*-4/5x

x=30

But then you are gonna want to plug 30 in for x in y=-4/5x-6 to make sure that it is the same as y=-30.

You might be interested in

Determine the equation of the parabola whose graph is given below. Write your answer in General Form. A parabola on a coordinate

dimaraw [331]

The equation of parabola becomes y = -2/25(x-3)^2 + 4.

According to the statement

we have given a graph and from this graph we have to find the equation of parabola in the general form.

So,

we know that the equation of parabola in general form is

y = a(x-h)^2 +k - (1)

From the graph we have:

a point on the graph is (x,y) = (-2,2)

the vertex of the graph is (h,k) = (3,4)

Now, substitute these values in the equation number (1)

Then

y = a(x-h)^2 +k

2 = a(-2-3)^2 +4

2 = a(-5)^2 +4

2 = a(25) +4

25a = -2

a = -2/25.

Now put a = -2/25 and (h,k) = (3,4) in the equation(1).

Then

the equation of parabola becomes y = -2/25(x-3)^2 + 4

So, The equation of parabola becomes y = -2/25(x-3)^2 + 4.

Learn more about equation of parabola here brainly.com/question/4061870

#SPJ4

6 0

1 year ago

Find a third-degree polynomial function with real coefficients and with zeros at -4 and -3 5i

Charra [1.4K]

If -3+5i is a solution, then by the conjugate root theorem, -3-5i is also a solution. We find the polynomial by multiplying together the factors. If x = -4, then x + 4 is the factor. If x = -3+5i, then (x-(-3+5i)) is a factor, and so is (x-(-3-5i)). Simplifying those down gives us as the first factor as (x+3-5i) and the second as (x+3+5i). We can FOIL those 2 together to get their product, and then FOIL in x+4. FOILing the 2 complex factors together gives us $x^2+3x+15ix+3x+9+15i-15ix-15i-25i^2$ . If we combine like terms and cross out things that cancel it's much easier than what it looks like there! It simplifies down to $x^2+6x-25i^2$ . Since i^2 = -1, it simplifies further to $x^2+6x-25(-1)$ and finally, to $x^2+6x+25$ . Now we will FOIL in x+4. $(x^2+6x+25)(x+4)=x^3+6x^2+25x+4x^2+24x+100$ . Our final simplified third degree polynomial is $x^3+10x^2+49x+100$