All categories

3 years ago

Andre and Diego were each trying to solve 2x+6=3x-8 Describe the first step they each make to the equation

Mathematics

1 answer:

avanturin [10]3 years ago

6 0

The first step was the combine the like terms; in this example he is subtracting the 3x on both sides of the =.

If you were to solve the equation it would look like this:

2x+6=3x-8
-3x -3x

-x+6=-8
-6 -6

-x=-14
/-1 /-1

x=14

You might be interested in

The non separable differential equation dy dx = 8x − 2y has a linear particular solution of the form y = mx + b. Find m + b.

yulyashka [42]

Answer:

$m + b=2$

Step-by-step explanation:

From the question we are told that

The non separable differential equation $\frac{dy}{dx} = 8x - 2y$

Generally the chi form of a give equation is mathematically represented as

$\frac{dy}{dx} +P(x)y= q(x)$

$\frac{dy}{dx} + 2y = 8x$

$ye^2^x=\int e^2^.8xdxx$

$ye^2^x=8(\frac{xe^2^x}{2}-\frac{\int e^2^x}{2})$

$ye^2^x=8(\frac{xe^2^x}{2}-\frac{e^2^x}{1})+c\\$

$ye^2^x=4x^2^x-2e^2^x+c$

$y=4x-2+c$

for c=0

$y=4x-2$

Therefore

$m=4\\b=-2\\m + b=4-(+2)$

$m + b=2$

6 0

3 years ago

30 points- help ASAP please.

Vera_Pavlovna [14]

Answer:

1) a) yes

b) no

c) a² - 39

(a)² - (sqrt(39))²

(a - sqrt(39))(a + sqrt(39))

This quadratic can be split into real factors, but not rational

sqrt(39) is a real number, but not rational

3 0

3 years ago

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.