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wlad13 [49]
1 year ago
7

Please please help Given the function f(x)=1/3x*2-3x+5 determine the inverse relation

Mathematics
1 answer:
yuradex [85]1 year ago
5 0

The inverse relation of the function f(x)=1/3x*2-3x+5 is f-1(x) = 9/2 + √(3x + 21/4)

<h3>How to determine the inverse relation?</h3>

The function is given as

f(x)=1/3x^2-3x+5

Start by rewriting the function in vertex form

f(x) = 1/3(x - 9/2)^2 -7/4

Rewrite the function as

y = 1/3(x - 9/2)^2 -7/4

Swap x and y

x = 1/3(y - 9/2)^2 -7/4

Add 7/4 to both sides

x + 7/4= 1/3(y - 9/2)^2

Multiply by 3

3x + 21/4= (y - 9/2)^2

Take the square roots

y - 9/2 = √(3x + 21/4)

This gives

y = 9/2 + √(3x + 21/4)

Hence, the inverse relation of the function f(x)=1/3x*2-3x+5 is f-1(x) = 9/2 + √(3x + 21/4)

Read more about inverse functions at:

brainly.com/question/14391067

#SPJ1

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qaws [65]

Answer:

2 x

— = —

25 700

700(2)= 25(x)

1400/25=x

x=56

Step-by-step explanation:

7 0
3 years ago
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Kendra had twice as much money as kareem. Kendra later spent $8 and kareem earned $6. By then,the two had the same amount of mon
Svet_ta [14]

Answer:

x = 2y

x - 8 = y + 6

replace x by 2y in the second equation

2y - 8 = y + 6

y - 8 = 6

y = 14

x = 2*14 = 28

Kendra had $28 and Kareem had $14.

6 0
3 years ago
Which input in this table is incorrect?
Elenna [48]

Answer:

A) 10.00

Step-by-step explanation:

10.00 x 0.8 is equal to 8 giving a total of 18.00. But it is incorrectly done  in the table.

6 0
2 years ago
Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
3 years ago
CAN SOMEONE PLEASE HELP ME ASAP ILL MARK BRAINLIST!!!!
slava [35]

Answer:

311/9

Step-by-step explanation:

trust me im big brain lol hope this helps

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