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Nezavi [6.7K]
3 years ago
15

The world population in 2013 was about 7.125 billion. In 2003, it was about 6.357 billion. Which number represents the increase

in the world population between 2003 and 2013?
Mathematics
2 answers:
murzikaleks [220]3 years ago
3 0
768 Im pretty sure. subtract 7.125-6.357
nexus9112 [7]3 years ago
3 0

Answer:

The number represent the increase in the world population is 0.768 billion.

Step-by-step explanation:

Given : The world population in 2013 was about 7.125 billion. In 2003, it was about 6.357 billion.

To find : Which number represents the increase in the world population between 2003 and 2013?

Solution :

The world population in 2013 was about 7.125 billion.

The world population in 2003 was about 6.357 billion.

The world population between 2003 and 2013 is

The population of 2013- The population of 2003

=7.125\text{ billion}- 6.357\text{ billion}

=0.768\text{ billion}

The number represent the increase in the world population is 0.768 billion.

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Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants. 37 x2 − 6x
Dvinal [7]

Answer:

\dfrac{A}{x}+\dfrac{B}{x-6}

Step-by-step explanation:

Given the function \dfrac{37}{x(x-6)}, to write the form of its partial fraction on decomposition, we will separate the two functions separated by an addition sign. The numerator of each function will be constants A and b and the denominator will be the individual factors of each function at the denominator. The partial fraction of the rational function is as shown below.

= \dfrac{37}{x(x-6)}\\\\= \dfrac{A}{x}+\dfrac{B}{x-6}

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8 0
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A ball is thrown up in the air, and it's height (in feet) as a function of time (in seconds) can be written as h(t) = -16t2 + 32
olga2289 [7]

Answer:

1) Using properties of the quadratic equation:

Here the height equation is:

h(t) = -16t^2 + 32t + 6

We can see that the leading coefficient is negative, this means that the arms of the graph will open downwards.

Then, the vertex of the quadratic equation will be the maximum.

Remember that for a general quadratic equation:

a*x^2 + b*x + c = y

the x-value of the vertex is:

x = -b/2a

Then in our case, the vertex is at:

t = -32/(2*-16) = 1

The maximum height will be the height equation evaluated in this time:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

2) Second method, using physics.

We know that an object reaches its maximum height when the velocity is equal to zero (the velocity equal to zero means that, at this point, the object stops going upwards).

If the height equation is:

h(t) = -16*t^2 + 32*t + 6

the velocity equation is the first derivation of h(t)

Remember that for a function f(x) = a*x^n

we have that:

df(x)/dx = n*a*x^(n-1)

Then:

v(t) = dh(t)/dt = 2*(-16)*t + 32 + 0

v(t) = -32*t + 32

Now we need to find the value of t such that the velocity is equal to zero:

v(t) = 0 = -32*t + 32

       32*t = 32

            t = 32/32 = 1

So the maximum height is at t = 1

(same as before)

Now we just need to evaluate the height equation in t = 1:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

4 0
3 years ago
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