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3 years ago

(NR) Natural Resources:Question 10

Mathematics

1 answer:

elena55 [62]3 years ago

5 0

Answer: D) Groundwater collects only where the soil and geological features allow water to seep down into underground chambers.

In other words, as long as there are gaps to allow water through, then that would lead to underground aquifers. Another important component is the amount of rain water that happens. Sources such as rivers, lakes and streams are also a contributing factor.

Stuff like earthquakes, volcanic activity, and continental movement doesn't really have much an effect on where or how aquifers form. It's really all about the soil type and how the landscape is formed, and of course the other factors mentioned in the previous paragraph.

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Explanation and answer please :)

Juli2301 [7.4K]

$20,000 is between $15,000 and $49,999, so we'll use the interest rate of 6.5% (see row 3)

r = 6.5% = 6.5/100 = 0.065

We'll use the decimal form of the interest rate as it is most common for financial math problems.

P = 20,000 is the amount deposited

t = 1 year is the amount of time

We will plug those values into the formula

i = P*r*t

to get the following:

i = P*r*t

i = 20000*0.065*1

i = 1300

So Mark earns $1,300 in simple interest each year.

6 0

3 years ago

Find all real solutions and show work: <br> 2x^3-3x^2+18x-27=0

Amiraneli [1.4K]

Given:

The given equation is

$2x^3-3x^2+18x-27=0$

To find:

All the real solutions.

Solution:

We have,

$2x^3-3x^2+18x-27=0$

It can be written as

$x^2(2x-3)+9(2x-3)=0$

$(2x-3)(x^2+9)=0$

Using zero product property, we get

$(2x-3)=0$ and $(x^2+9)=0$

$2x=3$ and $x^2=-9$

$x=\dfrac{3}{2}$ and $x=\pm\sqrt{-9}$

We know that $x=\pm\sqrt{-9}$ is an imaginary number because there is a negative sign under the square root.

Therefore, $x=\dfrac{3}{2}$ is the only real solution of the given equation.

5 0

3 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$