Solve the system below

Rewrite the equation 4x - 5y = 20 into slope-intercept form and graph it.

natta225 [31]

Y= 4 - 4/5x all you have to do is make it go up 4 from 0 and either go down 4 and go right 5 or up 4 and to the left 5... hope this helped!

6 0

3 years ago

Based on the graph below, what is yhe total number of solutions to the equation f(x)= g(x)?

valentinak56 [21]

Answer:

Based on the graph below, what is the total number of solutions to the equation f(x)= g(x) will be 3.

Step-by-step explanation:

The intersection points of both graphs would be the total number of solutions to the equation f(x)= g(x).

From the given diagram, it is clear that both the graphs intersect at three locations points or intersection points. The approximations locations of The intersection points of both graphs are

(1.5, 4.125)

(-1, 0), and

(-2.5, -3.5)

Therefore, based on the graph below, what is the total number of solutions to the equation f(x)= g(x) will be 3.

6 0

4 years ago

Graph the linear function f(x) =3x+2 and find the maximum on the interval [-1,1]

Gelneren [198K]

The purple dot is the interval (-1,1) and I don't think maximum and minimum exist on linear functions :/

4 0

3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .