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3 years ago

A line passes through the point (10,3)

1 answer:

5 0

We have to find b first:

3 = -3/2(10)+b
3 = -15+b
18 = b

Now right the equation, your slope will be the same one....

Y = -3/2x + 18

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the difference between two numbers is 342. the bigger number is 4 times the smaller number. find the product of the two numbers.

SVETLANKA909090 [29]

Let the two numbers be x and y.

x = 4y

x - y = 342

3y = 342

y = 114

xy = 4y * y = 456 * 114 = 51984

6 0

4 years ago

Please help me with this

kolbaska11 [484]

Answer:

Step-by-step explanation:

but its a negative 2 so but it would be 4 times 2 its 4 times negative 4

3 0

3 years ago

6-2x=5x-9x+10 solve for x

andrew-mc [135]

X=2

So you solve for x by simplifying bath sides of the equation, then all you do is isolate the variables

7 0

4 years ago

Determine which statements are true about the extrema of the graphed fourth degree polynomial. Select all that apply.

dusya [7]

C and D are the answers

7 0

3 years ago

Read 2 more answers

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