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4 years ago

If 35% of a natural area is to be developed, leaving 500 acres untouched, how many acres are to be developed?

2 answers:

8 0

Given that the 35% of area has been developed, the percentage of land that has not been developed is:
100-35
=65%
hence the acres of land that are to be developed will be:
35/65*500
=269.23 acres
the answer is 269.23

harkovskaia [24]4 years ago

5 0

Answer:

Area to be developed = 269.23 acres

Step-by-step explanation:

Let there is A acres natural area to be developed.

If 35% of A acres is developed then untouched area will be (100 - 35 = 65%)

Now it has been given in the question that 65% of A (untouched area) = 500 acres

$A.\frac{65}{100}=500$

$A=\frac{500}{0.65}=769.23$

So area of the natural field = 769.23 acres

Now 35% of the natural area which is to be developed = A×35%

$=769.23.\frac{35}{100}=(769.23)(.35)=269.23$

Area of the natural area to be developed is 269.23 acres

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Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$