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3 years ago

Help please!!!!!!!!!!!!!!!

Mathematics

2 answers:

IgorC [24]3 years ago

7 0

I believe the error was made in step 2

Ray Of Light [21]3 years ago

3 0

The Third One Is Correct

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What’s the quotient of 2 divided by 1/3

just olya [345]

2 4/5 divided by 1 1/3
= 14/5 divided by 4/3
= 14/5 * 3/4
= 42/20
= 21/10
= 2 1/10
Final Answer: 2 1/10

7 0

3 years ago

J+2 = 3k+5 in terms of k

leonid [27]

Answer:

(j-3)/3 =k

Step-by-step explanation:

j+2 = 3k+5

Solve for k

Subtract 5 from each side

j+2-5 = 3k+5-5

j-3 = 3k

Divide each side by 3

(j-3)/3 = 3k/3

(j-3)/3 =k

5 0

3 years ago

Please help ill mark brainliest

IgorLugansk [536]

Answer:

i think its 18

Step-by-step explanation:

5 0

3 years ago

What is factor of 4x+2y=

Simora [160]

2 is a factor of 4x + 2y
2( 2x + y)

8 0

3 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