Find the GCF of each set of numbers <br><br> 150, 200, and 300

Two students from each fourth and fifth grade class were on the planning committee.

Assoli18 [71]

Answer:

34

Step-by-step explanation:

17(2)=34

3 0

2 years ago

Which fraction correctly represents 0.54

boyakko [2]

Answer:

27/50 is the fraction.

Step-by-step explanation:

6 0

2 years ago

At Stephanie's new job, she spent $6.50, $8, $7.25, $13.50, and $9.75 on lunch the first week. In the second week, Stephanie spe

MaRussiya [10]

Answer:

The mean for the second week is $2 less than the first and in percentage it is 22% less.

Step-by-step explanation:

The mean is given by the sum of all individual values divided by the number of values. For the first week the sum is:

sum1 = 6.5 + 8 + 7.25 + 13.5 + 9.75

sum1 = 45

Since she spent 10 less in the second week the sum is:

sum2 = sum1 - 10 = 45 - 10 = 35

The mean for each week is:

mean1 = sum1/5 = 45/5 = 9

mean2 = sum2/5 = 35/5 = 7

difference = mean1 - mean2 = 9-7 = 2

difference(%) = [2/9]*100 = 0.22*100 = 22%

The mean for the second week is $2 less than the first and in percentage it is 22% less.

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

Please help me solve this, I will give you brainliest

ehidna [41]

2008 payan walang yan ha kaya mo yan

7 0

3 years ago

Find the GCF of each set of numbers 150, 200, and 300