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2 years ago

Whats two hundred fifty six thousand in standard form

Mathematics

2 answers:

Vikentia [17]2 years ago

8 0

256,000 .../////////////////

Gala2k [10]2 years ago

7 0

Standard form means write the normal way so the answer is 256,000

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1/8x8=? <br> As a mixed number and please explain

pishuonlain [190]

Answer:

1/8 · 8 = 1

Step-by-step explanation:

$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} = \frac{8}{8}$ or 1

hope this helps

4 0

2 years ago

Read 2 more answers

What does 256 to the power of 0 equal too?

morpeh [17]

Answer:

Step-by-step explanation:

Anything to the power of 0 is 1

Hope I helped :)

Please consider Brainliest :)

8 0

3 years ago

Woo-Jin and Kiran were asked to find an explicit formula for the sequence 64\,,\,16\,,\,4\,,\,1,...64,16,4,1,...64, comma, 16, c

olya-2409 [2.1K]

Answer:

$f(n)=64(\frac{1}{4})^{n-1}$

Step-by-step explanation:

The given sequence is

64, 16, 4, 1

$r_1=\dfrac{a_2}{a_1}=\dfrac{16}{64}=\dfrac{1}{4}$

$r_2=\dfrac{a_3}{a_2}=\dfrac{4}{16}=\dfrac{1}{4}$

$r_3=\dfrac{a_4}{a_3}=\dfrac{1}{4}$

It is a geometric series because it has a common ratio $r_1=r_2=r_3=\dfrac{1}{4}$ .

First term is 64.

The explicit formula of a geometric series is

$f(n)=ar^{n-1}$

where, a is first term and r is common ratio.

Substitute a=64 and r=1/4 in the above function.

$f(n)=64(\frac{1}{4})^{n-1}$

Therefore, the required explicit formula is $f(n)=64(\frac{1}{4})^{n-1}$ .

5 0

3 years ago

Read 2 more answers

Please explain and help me with this question.

k0ka [10]

it is so easy

Step-by-step explanation:

can i help you

5 0

2 years ago

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago