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VLD [36.1K]
3 years ago
14

What is the inverse of this function

Mathematics
1 answer:
just olya [345]3 years ago
6 0
G(x) = (25 - x)/5     First lets call g(x) by its other name - y

y = (25 - x)/5         Now switch the x and y

x = (25 - y)/5        Solve for y which is now the inverse - call this g^{-1}(x)

5x = 25 - y
y = 25 - 5x      this is the inverse so g^{-1}(x) = 25 - 5x
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5. Ana's age is twice the age of Lorna. If the sum of their ages is no more than 72, find the age of
ExtremeBDS [4]

Answer:

Ana:48

Lorna:24

Together they are a total of 72 yrs old

4 0
3 years ago
Read 2 more answers
Find the first three terms of f(n + 1) = f(n) + 8, if f(1) = 27.
dezoksy [38]
I hope this helps you



f (1+1)= f (1)+8


f (2)= 27+8


f (2)=35



f (2+1)= f (2)+8



f (3)= 35+8


f (3)=43



3 0
4 years ago
The cheapest available ticket for Super Bowl Ll on the secondary market costs
drek231 [11]

Answer:

He must work 52 days to pay for a single ticket.

Step-by-step explanation:

This question can be solved using proportions.

Per hour:

Joel earns $7.25 per hour, 20% of which is deducted for taxes. So without taxes, in each hour, he earns 100%-20% of 80% of this, so 0.8*7.25 = $5.8.

Per day:

He works 9 a.m. to 5 p.m. each day, so 8 hours a day.

For each hour, he earns $5.8.

So in a day, he makes 8*5.8 = $46.4

How many days he must work:

The ticket costs $2400.

He makes $46.4 a day.

So, to buy a ticket, he needs to work:

2400/46.4 = 51.7 days

Rounding up

He must work 52 days to pay for a single ticket.

7 0
3 years ago
PLEASE HELP WILL GIVE A BRAINLIEST!!
Tanya [424]
The answer is a. number of boys < 89.
7 0
3 years ago
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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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