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ratelena [41]
3 years ago
8

- You study 4 hours per day and are in class 6 hours

Mathematics
1 answer:
deff fn [24]3 years ago
7 0

Answer:

2/3

Step-by-step explanation:

Initially it would be 4/6 which you can simplify as 2/3

You might be interested in
Negative 4,200 divided by 300
Pachacha [2.7K]

Answer:

-14

Step-by-step explanation:

5 0
2 years ago
Is 68 prime because i am doing homework and im stuck on this and it is very hard so help me please
BlackZzzverrR [31]

If  68  is a prime number, then the only factors it has are  1  and  68.
If it has any other factors besides  1  and  68, then it's NOT prime.

Right away, without any higher math, you can look at just the last digit
in 68 .  The last digit is '8'.  That tells you that '68' is an even number,
and THAT tells you that '2' must be one of its factors.  So '68' is not a
prime number.

The factors of  68  are  1,  2,  4,  17,  34, and  68 .    

68  has four more factors besides  1  and  68, so it's not a prime number.

8 0
3 years ago
Read 2 more answers
I WILL GIVE BRAINLEST FOR THE CORRECT ANSWER!
pentagon [3]

Answer:

A) -4

B) 4 units apart

C) no values of x

Step-by-step explanation:

1) slope of m: (3-2)/(10-8) = ½

m(x) = ½x + c

2 = ½(8) + c

c = -2

m(x) = ½x - 2

m(16) = 6

h(4) = (4-2)²/2 = 4/2 = 2

h(4)-m(16) = 2-6 = -4

B) y-intercept of:

*h(x) = ½(0-2)² = 2

*m(x) = -2

from the equation found before

2 - (-2) = 4 units apart

C) m(x) = (x-4)/2

m(x) > h(x)

(x-4)/2 > ½(x-2)²

x-4 > x²-4x+4

x²-5x+8 < 0

(x - 2.5)² + 1.75 < 0

Never negative

3 0
3 years ago
there are 1,392 chairs in the auditorium. the seats are arranged 29 in each row. How many rows of 29 seats are there?
scoray [572]
There are 48 rows in the auditorium

7 0
3 years ago
Read 2 more answers
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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