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3 years ago

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in a set of 9 balls, 5 are blue. the balls are placed into a bag. mark is trying to get a ball that is not blue. he picks one ba

ll, puts it back, and then picks another ball. what is the probability he picked a blue ball both times

1 answer:

harina [27]3 years ago

4 0

So what you will have to do is 45 alla you have to do is multiple

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F f(x) = - 4x + 3 and g(x) = 3x2 + 2x - 4, find (f+ g)(x).

Alla [95]

Answer:

A.

${ \bf{(f + g)x = {(3x}^{2} + 2x - 1) + ( - 4x + 3) }} \\ = ( {3x}^{2} - 2x + 2)$

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3 years ago

Please help need answer

Len [333]

Answer:

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3 years ago

(2a+b)^2÷(3b-1) when a= -2 and b=5

WINSTONCH [101]

Evaluate (2 a + b)^2/(3 b - 1) where a = -2 and b = 5:
(2 a + b)^2/(3 b - 1) = (5 - 2×2)^2/(3×5 - 1)

3×5 = 15:
(5 - 2×2)^2/(15 - 1)

| 1 | 5
- | | 1
| 1 | 4:
(5 - 2×2)^2/14

2 (-2) = -4:
(-4 + 5)^2/14

5 - 4 = 1:
1^2/14

1^2 = 1:

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6 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