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podryga [215]
3 years ago
9

Finding the better estimate of

lt=" \sqrt{2} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Alenkinab [10]3 years ago
7 0

There's no such thing as a "better" (or best) estimate of the square root of 2. This is an irrational number, i.e. it has infinitely many decimal digits, and they don't follow any patter.

So, the most you can do is approximate the number to an arbitrary number of decimal digits. For example, if you want 4 decimal digits, you have

\sqrt{2}\approx 1.4142

You might be interested in
Factor x^2–a^2+3(x+a)
Mandarinka [93]
Answers
(x + a)^2 + 3(x + a) + 2
= [(x + a)(x + a + 3)] + 2
= (a + x + 2) (a + x + 1)
5 0
4 years ago
Elena and jada are going home. Elena runs at a constant speed of 4 miles per hour, and jada walks at a constant speed of 2 miles
BaLLatris [955]

Answer:

Elena will run 6 miles in 1.5 hours

Jada will walk 3 miles in 1.5 hours

Step-by-step explanation:

Multiply: 4*1.5=6 (Elena's)

Multiply:2*1.5=3  (Jada's)

Elena will run 6 miles in 1.5 hours

Jada will walk 3 miles in 1.5 hours

<em>Hope this helped!! :)</em>

<em>Stay safe and have a wonderful day/night!!!!!</em>

<em>Brainliest?!?!</em>

<em>Please correct me if I'm wrong</em>

5 0
3 years ago
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
Rearrange the formula to make the letter in the square bracket the subject <br> y= 3x - 5 (x)
patriot [66]

Answer:

y= 3x - 5

y+5 = 3x

x = (y+5)/3

8 0
3 years ago
Read 2 more answers
Help me! i have 20 minutes​
KIM [24]

Your on your own.. :/

5 0
3 years ago
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