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Dennis_Churaev [7]
3 years ago
15

PLEASE HELP ANSWER THESE QUESTIONS!

Mathematics
1 answer:
Sliva [168]3 years ago
7 0

Answer:

Step-by-step explanation:

a² - b² = (a+ b)(a - b)

1) (2n-4/2n) ÷ (n^2-4/n)

=\frac{2n-4}{2n}*\frac{n}{n^{2}-4}\\\\=\frac{2n-2*2}{2n}*\frac{n}{n^{2}-2^{2}}\\\\=\frac{2*(n-2)}{2n}*\frac{n}{(n+2)*(n-2)}\\\\=\frac{1}{n+2}

2) [y^2-36/y^2-49] ÷[ y+6/y-7]

=\frac{y^{2}-36}{y^{2}-49}*\frac{y-7}{y+6}\\\\=\frac{y^{2}-6^{2}}{y^{2}-7^{2}}*\frac{y-7}{y+6}\\\\=\frac{(y+6)*(y-6)}{(y+7)*(y-7)}*\frac{y-7}{y+6}\\\\=\frac{y-6}{y+7}\\

3) [m^2-1/ m^2-m] ÷ [m^2-7m-8/3m ]

=\frac{m^{2}-1}{m^{2}-m}*\frac{3m}{m^{2}-7m-8}\\\\=\frac{(m+1)*(m-1)}{m*(m-1)}*\frac{3m}{(m-8)*(m+1)}\\\\=\frac{3}{m-8}

Hint :  m² - 7m - 8

sum = -7

Product  = -8

Factor = (-8), 1

m² - 7m - 8  =m² - 8m + m - 8

                   = m*(m - 8) + (m-8)

                   = (m - 8)(m +1)

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<h2>Half Life</h2>

The half life period is the time in which only half of the given population remains. It can be represented through this equation:

f(t)=a\times(1/2)^{\frac{t}{h}}

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  • <em>h</em> = half life

<h2>Solving the Question</h2>

We're given:

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  • <em>a</em> = 184 grams (this is the initial mass, after 0 time has passed)

For most questions like this, we would have to plug these values into the equation mentioned above. However, this question asks for the time elapsed after 3 half-lives.

This can be calculated simply by multiplying the given half-life by 3:

28 million years x 3

= 84 million years

<h2>Answer</h2>

84 million years

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I need to know the answer to number one
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A gun mass of 5 kg fired a bullet of mass 10 g with the velocity of 360 km/h. What
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We also write the mass of the bullet in the same units as the mass of the gun (for example kilograms). Mass of the bullet = 0.010 kg

In mathematical terms, we have:

5\, kg * v= 0.01 \,kg\,* 360\,\frac{km}{h} \\v=\frac{0.01\,*360}{5} \,\,\frac{km}{h}\\v=0.72\,\,\frac{km}{h}

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Answer: 2,160\ ft^3

Step-by-step explanation:

The first step is to find the ratio of the lengths.

According to the information given in the exercise, one the solids has edges of  12 feet and the other solid has edges of 24 feet.

Therefore, the ratio of the length of the smaller solid to the length of the is the following:

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Now, the ratio to the volumes of the smaller solid to the other one is the following:

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