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3 years ago

Which fractions are greater than 5/10 ? Choose exactly two answers that are correct.

Mathematics

2 answers:

lakkis [162]3 years ago

8 0

6/10 or 3/5
7/10
What are the choices for the answer?

WINSTONCH [101]3 years ago

4 0

Hey there!

Question: what fractions are greater than $\frac{5}{10}$ ? Choose exactly 2 answers

Well honestly, we can find our answer by converting the answers to decimals

$\frac{5}{10} = \frac{1}{2} = 0.5 \\ \\ \\ \\ \frac{6}{12} = \frac{1}{2} = 0.5 \\ \\ \frac{2}{3} = \frac{2}{3} = 0.667 \\ \\ \frac{63}{100} = \frac{63}{100} = 0.63 \\ \\ \frac{3}{8} = \frac{3}{8} = 0.375$

It can't be A because it is equivalent to the fraction

It can be B because it is greater than the fraction

It can be C because it is also greater than the fraction

It can't be D because it is lesser than the fraction

Answer: $B,C$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

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

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