All categories

2 years ago

1. suppose you purchased a new car for

Mathematics

1 answer:

Xelga [282]2 years ago

8 0

If the value of the car decreases by 8% every year it would take 13 years for the car to be worth
$10000.00.

You might be interested in

Paz has $18 in her wallet. This is 3 times the money in her pocket. Write and solve an equation to find out how much money is in

taurus [48]

6. the answer is 6. because 3x what she has is equal to 18 all u do is divide. u get six.

6 0

3 years ago

suppose a charity received a donation of $22.8 million. if this represents 58% of the charity's donated fund, what is the total

Mrac [35]

The total amount is "x", so, if 22.8 is 58%, "x" is the 100%,

$\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
22.8&58\\
x&100
\end{array}\implies \cfrac{22.8}{x}=\cfrac{58}{100}\implies \cfrac{22.8\cdot 100}{58}=x$

6 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago

I need help on this question

wel

Answer:

The attachment is black

Step-by-step explanation:

5 0

3 years ago

The length of a rectangular garden is 5 meters long. The area of the garden is 10 square meters. Which equation correctly relate

serious [3.7K]

Answer:

the width is 2 meters

Step-by-step explanation:

10 / 5 = 2

3 0

3 years ago