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3 years ago

3. Every time Deucalion threw one stone over his shoulder, eight people sprang

Mathematics

1 answer:

kvv77 [185]3 years ago

6 0

Answer:

1156

Step-by-step explanation:

When Deucalion threw one stone over his shoulder, 8 people appear. So, for every stone, we have 8 new people.

Therefore, if Deucalion threw 68 stones, the number of people that would appear is

$n_1=68\cdot 8 =544$

Instead, when Phyrra threw a stone, 9 people appear. So, for every stone, we have 9 new people.

Therefore, if Phyrra threw 68 stones, the number of people that would appear is

$n_2=68\cdot 9 =612$

And so, we can now calculate the total number of people that appear after each had thrown 68 stones:

$n=n_1+n_2=544+612=1156$

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Evaluate the expression 9P1

garri49 [273]

Answer :

Explanation :

The expression above signifies the Permutation expression

Permutation expression is given as nPr

The Permutation expression is calculated as:

nPr = P(n, r) = n! / (n - r)!

It the above question, we were asked to evaluate the expression 9P1

From the above question, we can see that:

n = 9 and r = 1

Therefore, the expression

9P1 = 9! / ( 9 - 1) !

9P1 = 9! / 8!

= 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 / 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 362880 /40320

= 9

Therefore, the evaluation of the expression 9P1 = 9

4 0

3 years ago

Finds the solutions to x^2=20

ELEN [110]

Answer:

the solution of:

x²=20

√x² = √(2² × 5)

x=2√5

So the answer is

A. x= ±2√5

6 0

3 years ago

1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$

$\mathrm{Cancel\:the\:common\:factor:}\:10^{-12}$

$=\frac{12.6}{35.4\pi \left(10^{-2}\times \:2.5\right)^2}$

$=\frac{12.6}{0.025^2\times \:35.4\pi }$ ∵ $\left(10^{-2}\times \:2.5\right)^2=0.025^2$

$=\frac{12.6}{0.022125\pi }$ ∵ $35.4\pi\times 0.025^2=0.022125\pi$

$=\frac{12.6}{0.06950 }$

F $=181.274$ N

Therefore, the magnitude of the electric force between these two objects will be: 181.274 N.

i.e. F $=181.274$ N

7 0

3 years ago

The amount of time is takes dora to commute to work is always within 5 minutes of the average number of minutes the commute take

Maslowich

Answer:

The average time to commute to work is 25 minutes.

Step-by-step explanation:

If average time for Dora to commute to work is 5 mins. If she takes 45 mins to get to work then her overall average time to commute to work will change as the average consists of 2 or more figures. The average time to commute to work is 25 mins ((5+45)/2). The average of 5 mins and 45 mins is 25 mins. This is the overall average time she takes to commute to her work.

5 0

3 years ago

Find a polynomial that describes the area of the lawn which kevin has to mow in his front yard 6x+5,4x-9

Likurg_2 [28]

Answer:

$Area=24x^2-34x-45$