A balance can measure up to 0.001 g.when using such a balance, how many decimal places should your answer have?

Need help in this mathquestion. The function h(x)=(x+2)^2 can be expressed in the form f(g(x)) where f(x)=x^2 , and g(x) is defi

blsea [12.9K]

Answer:

In a nutshell, $g(x) = x+2$ .

Step-by-step explanation:

A composition is operation between function that consist in replacing the independent variable of the first function, $f(x)$ in this case, for $g(x)$ . Common notations for composition between functions are $f(g(x))$ and $f\circ g (x)$ . If we know that $h(x) = f\circ g (x) = (x+2)^{2}$ and $f(x) = x^{2}$ , then $g(x) = x+2$ .

In a nutshell, $g(x) = x+2$ .

4 0

3 years ago

Use integration by parts to find the integrals in Exercise. ∫(4x-12)e-8x dx.

stepladder [879]

Answer:

$e(2x^{2} -12x)-4x^{2}+C$

Step-by-step explanation:

We have been given an indefinite integral as $\int \left(4x-12\right)e-8x\:dx$ . We are asked to find the given integral.

Let us solve our given problem.

$\int \left(4x-12\right)e\:dx-\int 8x\:dx$

Take out constant:

$e\int \left(4x-12\right)\:dx-8\int x\:dx$

$e(\int 4x\:dx -\int 12\right\:dx)-8\int x\:dx$

$e(\frac{4x^{1+1}}{2} -12x)-8*\frac{x^{1+1}}{1+1}+C$

$e(\frac{4x^{2}}{2} -12x)-8*\frac{x^{2}}{2}+C$

$e(2x^{2} -12x)-4x^{2}+C$

Therefore, our required integral would be $e(2x^{2} -12x)-4x^{2}+C$ .

5 0

3 years ago

Find k so that the following function is continuous: f(x)={kx8x2if0≤x<5if5≤x.

tankabanditka [31]

Check the one-sided limits:

$\displaystyle \lim_{x\to5^-}f(x) = \lim_{x\to5}kx = 5k$

$\displaystyle \lim_{x\to5^+}f(x) = \lim_{x\to5}8x^2 = 200$

If f(x) is to be continuous at x = 5, then these two limits should have the same value, which means

5k = 200

k = 200/5

k = 40

3 0

3 years ago

write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

Liula [17]

We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.