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Marianna [84]
3 years ago
15

What is the value of the expression? −1/4+3/8 5/8 Enter your answer in simplest form.

Mathematics
1 answer:
galina1969 [7]3 years ago
7 0
Decimal form
-0.015625

Exact form
-1/60
Negative one over sixty
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I NEED HELP WITH A PROBLEM ​
postnew [5]

Answer:

the first one I guess

Step-by-step explanation:

6 0
3 years ago
What is the price of the items Nakita bought?
timurjin [86]

Answer:

B $10.45

Step-by-step explanation:

Fist you find how much the 2 boxes of cracker cost

$3.50 * 2 = $7

Then subtract the amount of 2 crackers by the $0.80 coupon

$7 - $0.80 = $6.20

Then add that amount to the cost of a jar of peanut butter

$6.20 + $4.85 = $11.05

Then add that about for with the amount of a package of juice boxes

$11.05 + $2.40 = $13.45

Then subtract the total amount to the $3.00 coupon

$13.45 - $3.00 = $10.45

B $10.45 is your answer

4 0
3 years ago
Determine the nth term for each of the following sequences:16,15,14,13,12,..
Vladimir79 [104]
F(n) = 1st term - common difference (n - 1)

The operation is subtraction because the sequence is decreasing.

n = number of the term you are looking for
1st term = 1st number in the sequence, in this case, 16
common difference = difference between one number to the next

f(n) = 16 - 1(n-1)

Assuming we are looking for the 5th term.
f(5) = 16 - 1(5-1)
       = 16 - 1(4)
       = 16 - 4
       = 12     * as you can see, 12 is the 5th term of the sequence.

If you are looking for the equivalent of the nth term, simply substitute n by the number you are looking for and solve the equation.
7 0
3 years ago
A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min
Dmitry [639]

At any time t (min), the volume of solution in the tank is

500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}

If A(t) is the amount of salt in the tank at any time t, then the solution has a concentration of \dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}.

The net rate of change of the amount of salt in the solution, A'(t), is the difference between the amount flowing in and the amount getting pumped out:

A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)

Dropping the units and simplifying, we get the linear ODE

A'=10+5\sin\dfrac t4-\dfrac A{10}

10A'+A=100+50\sin\dfrac t4

Multiplying both sides by e^{10t} allows us to identify the left side as a derivative of a product:

10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}

\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}

e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt

Integrate and divide both sides by e^{10t} to get

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}

The tanks starts off with 30 lb of salt, so A(0)=30 and we can solve for C to get a particular solution of

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}

6 0
3 years ago
5. Mr. Young bought a laptop for his small business for $ 1599. The value of the laptop depreciates at a rate of 19% each year U
stepladder [879]
The initial cost over 4 years?? What are you talking about
8 0
3 years ago
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