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2 years ago

If f(x) = x^2 +4, find f'(1)

Mathematics

1 answer:

marissa [1.9K]2 years ago

5 0

Answer:

f'(1x) = 2x

f'(0) = 4

Step-by-step explanation:

Hope this helps if not im sorry

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Write an equation in slope intercept form for the line passing through (2, 6) with a slope of -3

mixer [17]

Answer:

y=-3x+12

Step-by-step explanation:

y-y1=m(x-x1)

y-6=-3(x-2)

y-6=-3x+6

y=-3x+6+6

y=-3x+12

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2 years ago

Simplify ---<br>(2√7)^2

PolarNik [594]

$(2 \sqrt{ {7} } )^{2} = {2 }^{2} \times 7 = 4 \times 7 = 28$

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3 years ago

3/5x + 3 = 6 whats is x?

KengaRu [80]

Answer:

3/5x + 3 = 6

3/5x = 6 - 5

3/5x = 1

<h3><u>Multiplying </u><u>both </u><u>the </u><u>sides </u><u>with </u><u>5x </u></h3>

3/5x (5x) = 1* 5x

3 = 5x

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2 years ago

Which equation is a point slope form equation for line AB ?

zepelin [54]

(-2,3)(4,-1)
slope = (3 + 1)/(-2 - 4) = 4/-6 = -2/3

equation
y + 1 = -2/3(x - 4)

7 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