You would transform 5x + 2x into the same term with the associative property, adding like terms. If you subsitute the different values into the expressions you notice that both have a rate of change of 7 like an arithmetic sequence. To see if any value will make the two expressions equal you make them equal to each other.

5x + 2x = 7x -1

7x = 7x - 1

0 = -1

No they will never have be a solution to both expressions to make them equal because they're parallel to each other.

7 0

3 years ago

Complete the point slope equation of the line through (-5, 7) and (-4, 0).

IgorLugansk [536]

For this case we have that by definition, the point-slope equation of a line is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

We have two points:

$(x_ {1}, y_ {1}): (- 5,7)\\(x_ {2}, y_ {2}): (- 4,0)$

We found the slope:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0-7} {- 4 - (- 5)} = \frac {-7} {-4 + 5} = \frac {-7} {1} = - 7$

Thus, the equation is of the form:

$y = -7x + b$

We substitute one of the points and find "b":

$0 = -7 (-4) + b\\0 = 28 + b\\b = -28$

Finally, the equation is:

$y = -7x-28$

Answer:

$y = -7x-28$

5 0

3 years ago

Help help math math math

natka813 [3]

Answer:

Yes, this is a function

Step-by-step explanation:

You know this because it passes the vertical line test.

I hope this helps!

7 0

1 year ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago