All categories

3 years ago

Find the lowest common denominator.

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

4 0

The LCD is $(x+2) ^{2}(x-2) ^{2}$ since $x^{2} -4$ factors into (x+2)(x-2). The idea for a LCD is to have in all the denominators what makes up each individual denominator.

You might be interested in

WHAT IS THE EQUATION ?? AND TELL ME HOW YOU GOT IT.

tigry1 [53]

Answer:

SUM OF ALL ANGLES OF A TRIANGLE =180

THEREFORE,

X+8+2X-3+6X-5=180

9X=180

X=20.

THEREFORE,<J=X+8=20+8=28

<K=6X-5=6*20-5=120-5=115

<L=2X-3=2*20-3=40-3=37.

6 0

3 years ago

Differentiate the following equation

Bingel [31]

Answer:

$\frac{dy}{dx} = \frac{4}{1-x}$

Step-by-step explanation:

y = 3 - 4ln(1 - x)

⇒ $\frac{dy}{dx} =3\frac{d}{dx} -4ln(1-x)\frac{d}{dx}$

⇒ $\frac{dy}{dx} =0 -4ln(1-x)\frac{d}{dx}$

⇒ $\frac{dy}{dx} =0 -4(\frac{1}{1-x})\times-1$

⇒ $\frac{dy}{dx} = \frac{4}{1-x}$

4 0

2 years ago

The three straight paths, , , and , meet each other at three points, A, B, and C. How do these points of intersection differ fro

stiv31 [10]

Answer:

Step-by-step explanation:

I am sort. I have no idea

5 0

2 years ago

What is 3 to the power of 3 over 2 equal to?

ddd [48]

27/2. 3x3x3 is 27. .

5 0

3 years ago

Read 2 more answers

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago