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3 years ago

What is the simple interest accumulated on $150 at 3% for 6 years?

Mathematics

2 answers:

juin [17]3 years ago

8 0

150x6=900 that will be the answer

kondaur [170]3 years ago

6 0

The answer is 900 which you would get by multiplying 150 * 6 = 900 I hope this helped!!

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The dimensions of a rectangle are x+3 and x+4. The area of the rectangle is 72. Set up an equation that models this rectangle an

vodomira [7]

Answer: x = 5

Step-by-step explanation:

The area of a rectangle is the length times the width. Since the dimensions are given as functions of x, youre going to have to use FOIL to combine them, then set that answer equal to the given value of 72.

l*w = (x+3)*(x+4) = x^2 + 7x +12 = 72

Subtract 72 on both sides. There is a reason why we'll do this instead of subtracting 12. The quadratic formula will help us to solve for x.

x^2 + 7x - 60 = 0

Apply the quadratic formula

$\frac{-7 (+/-)\sqrt{7^2-(4)*(1)*(-60} }{2*1} =\frac{-7(+/-)17}{2}$

We cant have a negative x value because that would result in negative dimensions and a negative area, which is impossible in this context. So you want to add the 17 instead of subtracting it.

x = 5 and x = -12

An alternate way is just factoring the trinomial to:

(x+12)(x-5) = 0

3 0

3 years ago

A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) Large and small drains are opened

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) Only the small drain is opened

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) Only the big drain is opened

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$