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2 years ago

Andrea gave 1 widget and received 3; Brian gave 3 widgets and received 13; and Chris forgot how many widgets he gave,

Mathematics

1 answer:

Bess [88]2 years ago

6 0

Answer:

Chris gave 4 and got 28

Dan gave 4 and got 28

this is the best I can do

sorry

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Jared and Morgan are making two gazebos in the shape of regular polygons, Jared's gazebo is a heptagon

bezimeni [28]

Answer:

10.4 :)

Step-by-step explanation:

6 0

3 years ago

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In a hostel, 80 students have food enough for 45 days. How many students should leave the hostel so that the food is enough for

Aleonysh [2.5K]

Say that taking a quarter of student out of the group will add a quarter present of days for food so by take 20 students out you’ll get 60 days (that’s what I got hope this helps)

4 0

3 years ago

Question 7 (Essay Worth 5 points)

kogti [31]

Answer:

x = -2

y=-3

(-2,-3)

Step-by-step explanation:

Both equations are equal to y

We can set them equal to each other

y = 3x + 3

y = x − 1

3x+3 = x-1

Subtract x from each side

3x+3 -x = x-1-x

2x+3 = -1

Subtract 3 from each side

2x+3-3 = -1-3

2x = -4

Divide each side by 2

2x/2 = -4/2

x = -2

Now we need to find y

y = x-1

y = -2-1

y = -3

y = x-1

3 0

3 years ago

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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

Find the percent of change the original cost was 50. And the new cost is 60

Ivan

Answer:

20%

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers