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3 years ago

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In 1940, a certain country's population was 79,042,573 and its area was 2,919,166 square miles. In 2000, the country's popul

ation was 283,891,555 and its area was 3,678,844 square miles. Compute the population densities of both years.

1 answer:

Debora [2.8K]3 years ago

3 0

Answer:

Step-by-step explanation:

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Find the optimal solution for the following problem. (Round your answers to 3 decimal places.)

Sphinxa [80]

Answer:

x=2.125

y=0

C=19.125

Step-by-step explanation:

To solve this problem we can use a graphical method, we start first noticing the restrictions $x\geq 0$ and $y\geq 0$ , which restricts the solution to be in the positive quadrant. Then we plot the first restriction $8x+10y\leq 17$ shown in purple, then we can plot the second one $11x+12y\leq 25$ shown in the second plot in green.

The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.

So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.

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3 years ago

1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

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2 years ago

Find the solutions to x² = 12.

Novosadov [1.4K]

Answer:

$x = \pm 2\sqrt3$

Step-by-step explanation:

Hello!

Let's use the square root property to solve this question

Solve:

$x^2 = 12$
$\sqrt{x^2} = \sqrt{12}$
$x = \pm\sqrt{12}$
$x = \pm \sqrt{4 * 3}$
$x = \pm 2\sqrt3$

The answer is option A: x = ±2√3

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2 years ago