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2 years ago

Mathematics with applications in the management, Natural, and Social Sciences Twelfth edition

Mathematics

1 answer:

Monica [59]2 years ago

4 0

Answer:

In 2015 the both populations were the same and from that year the population of millennials surpassed the population of boomers

Step-by-step explanation:

$\left \{ {{10x+13y = 1125} \atop {-2x+7y = 495}} \right.$

x=14

Boomer: 10(14)+13y=1125

140+13y=1125

13y=1125-140

y= 985/13

y= 75.77 (75.77 millions of boomers in 2014)

Millenials: -2(14) +7y = 495

-28 +7y = 495

7y= 495+28

y= 523/7

y=74.71 (74.71 millios on millenials in 2014)

In 2014 the population of boombers were still greater than the population of millennials

The solution of the system of equations will give us the point where the populations were equalized, and from that point the population of boombers will be less than that of the millennials.

Boomers: 10x+13y = 1125

y= (-10x +1125)/13

Millenials: -2x+7y = 495

y= (2x+495)/7

We match both expressions of "y"

(-10x +1125)/13 =(2x+495)/7

cross multiply:

(-10x +1125)*7 =(2x+495)*13

-70x + 7875 = 26x + 6435

we group similar terms:

7875 -6435 = 26x+70x

1440 = 96x

x= 1440/96

x= 15

In 2015 the both populations were the same and from that year the population of millennials surpassed the population of boomers

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2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.