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3 years ago

How would the expression x^2+27 be rewritten using sum of cubes?

1 answer:

6 0

Assuming the square is not a typo, one can write

$x^2+27=(x^{2/3})^3+3^3=(x^{2/3}+3)\bigg((x^{2/3})^2-3x^{2/3}+9\bigg)$

$=(x^{2/3}+3)\bigg(x^{4/3}-3x^{2/3}+9\bigg)$

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4.76 to one decimal place

stellarik [79]

<span>The 6 means we round the 7 up to the next number which is 8. Thus, we round 4.76 to </span>4.8

3 0

3 years ago

Read 2 more answers

The midpoint of AB is M (2,0). If the coordinates of A are, (-3, 3), what are the coordinates of B?

gregori [183]

Given:

M is the midpoint of AB.

M(2,0) and A(-3, 3).

To find:

The coordinates of point B.

Solution:

Midpoint formula:

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Let the coordinates of point B are (a,b). Then, using the midpoint formula, we get

$(2,0)=\left(\dfrac{-3+a}{2},\dfrac{3+b}{2}\right)$

On comparing both sides, we get

$\dfrac{-3+a}{2}=2$

$-3+a=2\times 2$

$a=4+3$

$a=7$

And,

$\dfrac{3+b}{2}=0$

$3+b=0$

$3+b-3=0-3$

$b=-3$

Therefore, the coordinates of point B are (7,-3).

5 0

2 years ago

Find g(x) where g(x) is the translation 3 units right of f(x)=x^2

Fed [463]

Answer:

g(x) = x^2 -6x +9

Step-by-step explanation:

A function f(x) translated right h units and up k units will become ...

g(x) = f(x -h) +k

You want the function f(x) = x^2 to be translated right h=3 units, so it will become ...

g(x) = f(x -3) = (x -3)^2

g(x) = x^2 -6x +9

6 0

2 years ago

Type the correct answer in each box.

Leokris [45]

1 pound= $0.50
10 pounds= $5.00

3 0

3 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago