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Irina-Kira [14]
3 years ago
8

Help...me...plz..!!.....?

Mathematics
2 answers:
Montano1993 [528]3 years ago
8 0

Answer:

\huge\boxed{-1}

Step-by-step explanation:

\frac{-5^2}{(-5)^2}

=> \frac{-25}{25}

=> -1

Llana [10]3 years ago
8 0

Answer:

-1

Step-by-step explanation:

-5^2 is -25 because you have to do exponents first, so 5^2 is 25. Add a negative symbol there, it becomes -25.

For the denominator, we do the exact same process, except a little different. Since there are parentheses around the negative five, we have to apply the negative symbol on the 5 before we do the exponents. After, we do -5^2 which is 25 because when you are multiplying negative, you always get a positive.

When we are all done with that, we put the numerator over the denominator to get -25/25, which is equivalent to -1.

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The Edwards family's bill for dinner is $52.95. How much should they tip their server? (Assuming the normal 15%)
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Read 2 more answers
in circle T with m angle STU=50 and ST=3 units, find the length of arc SU. Round to the nearest hundredth.
ICE Princess25 [194]

9514 1404 393

Answer:

  arc SU ≈ 2.62 . . . units

Step-by-step explanation:

The arc length is given by ...

  s = rθ

where r is the radius, and θ is the central angle in radians.

  arc SU = ST·STU = 3·(50°·π/180°) = 5π/6

  arc SU ≈ 2.62 . . . units

6 0
3 years ago
On a class trip to the zoo, there is 1 chaperone for every 4 students. If 5 chaperones are on the trip, how many students are go
aleksandr82 [10.1K]

Answer:

20 Students

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6 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
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