Jubal wrote the four equations below. He examined them, without solving them, to determine which equation has no solution.

You need 150g of flour for 12 cakes, how much flour do you need for 30 cakes?

nirvana33 [79]

If you need 150g of flour for 12 cakes, that means you need 12.5 g of flour per cake. So if you multiply that by 30 it gives you
375g of flour

3 0

3 years ago

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5.) Divide 44,100 by 900 =

Aleonysh [2.5K]

Answer:

49

Step-by-step explanation:

44100/900=49

7 0

3 years ago

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Simplify the expression to a polynomial in standard form: (4x+5) (x^2-3x-2) PLEASE HLEP

pickupchik [31]

Answer:

$4x^3-7x^2-23x-10$

Step-by-step explanation:

So we have the expression:

$(4x+5)(x^2-3x-2)$

Distribute. Multiply (4x+5) to each term:

$(x^2)(4x+5)+(-3x)(4x+5)+(-2)(4x+5)$

Distribute:

$=(4x^3+5x^2)+(-12x^2-15x)+(-8x-10)$

Combine like terms:

$=(4x^3)+(5x^2-12x^2)+(-15x-8x)+(-10)$

Add and simplify:

$=4x^3-7x^2-23x-10$

And this is in descending order, so it's in standard form.

7 0

2 years ago

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.

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