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2 years ago

29/3 into a mixed fraction

Mathematics

1 answer:

garri49 [273]2 years ago

6 0

Answer: 29/3 into a mixed fraction would be 9 2/3

I hope I helped you. Have a great day!

Step-by-step explanation:

Simplify:

29/3 = 9 2/3

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Ming throws a stone off a bridge into a river below.

AURORKA [14]

Answer:

1 seconds after being thrown, the stone reaches its max height

Step-by-step explanation:

The parabolic (quadratic) equation is:

$h(x)=-5(x-1)^2+45$

Lets expand this in the form $ax^2+bx+c$ , so we have:

$h(x)=-5(x-1)^2+45\\h(x)=-5(x^2-2x+1)+45\\h(x)=-5x^2+10x-5+45\\h(x)=-5x^2+10x+40$

We can say the values of a,b, and c, now to be:

a = -5

b = 10

c = 40

The number of seconds at which the max would occur is given by the point, x, at:

$x=-\frac{b}{2a}$

We know a and b, let's find the seconds, x,

$x=-\frac{b}{2a}=-\frac{10}{2(-5)}=-\frac{10}{-10}=--1=1$

Hence,

1 seconds after being thrown, the stone reach its max height

6 0

3 years ago

Read 2 more answers

Custom Office makes a line of executive desks. It is estimated that the total cost for making x units of their Senior Executive

Ivan

Answer:

(a) The average cost function is $\bar{C}(x)=95+\frac{230000}{x}$

(b) The marginal average cost function is $\bar{C}'(x)=-\frac{230000}{x^2}$

(c) The average cost approaches to 95 if the production level is very high.

Step-by-step explanation:

(a) Suppose $C(x)$ is a total cost function. Then the average cost function, denoted by $\bar{C}(x)$ , is

$\frac{C(x)}{x}$

We know that the total cost for making x units of their Senior Executive model is given by the function

$C(x) = 95x + 230000$

The average cost function is

$\bar{C}(x)=\frac{C(x)}{x}=\frac{95x + 230000}{x} \\\bar{C}(x)=95+\frac{230000}{x}$

(b) The derivative $\bar{C}'(x)$ of the average cost function, called the marginal average cost function, measures the rate of change of the average cost function with respect to the number of units produced.

The marginal average cost function is

$\bar{C}'(x)=\frac{d}{dx}\left(95+\frac{230000}{x}\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g\\\\\frac{d}{dx}\left(95\right)+\frac{d}{dx}\left(\frac{230000}{x}\right)\\\\\bar{C}'(x)=-\frac{230000}{x^2}$

(c) The average cost approaches to 95 if the production level is very high.

$\lim_{x \to \infty} (\bar{C}(x))=\lim_{x \to \infty} (95+\frac{230000}{x})\\\\\lim _{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim _{x\to a}f\left(x\right)\pm \lim _{x\to a}g\left(x\right)\\\\=\lim _{x\to \infty \:}\left(95\right)+\lim _{x\to \infty \:}\left(\frac{230000}{x}\right)\\\\\lim _{x\to a}c=c\\\lim _{x\to \infty \:}\left(95\right)=95\\\\\mathrm{Apply\:Infinity\:Property:}\:\lim _{x\to \infty }\left(\frac{c}{x^a}\right)=0\\\lim_{x \to \infty} (\frac{230000}{x} )=0$