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2 years ago

What is the inverse of the function f(x) = one-quarterx – 12? h(x) = 48x – 4 h(x) = 48x + 4 h(x) = 4x – 48 h(x) = 4x + 48

Mathematics

2 answers:

DaniilM [7]2 years ago

8 0

Answer:

h(x) = 4x+48

Step-by-step explanation:

Given the function f(x) = 1/4x -12. To get the inverse of the function, first we will let y = f(x) to have y = 1/4x - 12.

Then we will make x the subject of the formula as shown:

y = 1/4x - 12

adding 12 to both sides we have;

y + 12= 1/4x - 12 + 12

y + 12 = 1/4x

Multiplying both sides by 4 we have;

4(y+12) = x

x = 4y + 48

Substituting y for x

y = 4x+48

f^-1(x) = 4x+48

h(x) = 4x+48

Thee final expression will give the inverse of the function

Basile [38]2 years ago

4 0

Answer:

Step-by-step explanation:

Given the function f(x) = 1/4x -12. To get the inverse of the function, first we will let y = f(x) to have y = 1/4x - 12.

Then we will make x the subject of the formula as shown:

y = 1/4x - 12

adding 12 to both sides we have;

y + 12= 1/4x - 12 + 12

y + 12 = 1/4x

Multiplying both sides by 4 we have;

4(y+12) = x

x = 4y + 48

Substituting y for x

y = 4x+48

f^-1(x) = 4x+48

h(x) = 4x+48

Thee final expression will give the inverse of the function

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Answer:

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Step-by-step explanation:

From the question we are told that

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Generally the probability that the selected pool has a pH level that is not considered safe is mathematically represented as

$P(A) = 1 - P(7.2 \le X \le 7.8 )$

Here

$P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - \mu }{\sigma } < \frac{X - \mu }{ \sigma }$

Generally $\frac{X - \mu }{ \sigma } = Z (The \ standardized \ value \ of X )$

$P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - 7.5 }{0.2 } < Z$

$P(7.2 < X < 7.8 ) = P(-1.5 < Z$

=> $P(7.2 < X < 7.8 ) = P(Z < 1.5) - P( Z < - 1.5)$

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$P(Z < 1.5) = 0.93319$

and

$P(Z < -1.5) = 0.066807$

$P(7.2 < X < 7.8 ) =0.93319 - 0.066807$

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2 years ago

Find the cube roots of 27(cos 330Á + i sin 330Á). Please help c:

Mashcka [7]

Use formula
$e^{ix}=\cos{x}+\sin{x}$

$\sqrt[3]{27(\cos{330^o}+i\sin{330^o})} 
\\\sqrt[3]{27e^{i330^o} }
\\ \sqrt[3]{27} \sqrt[3]{e^{i330^o}} 
\\ \sqrt[3]{3^3} \sqrt[3]{e^{i110^o\times 3}} 
\\ \sqrt[3]{3^3} \sqrt[3]{(e^{i110^o})^ 3}} 
\\ 3e^{i110^o}
\\3(\cos{110^o}+i\sin{110^o})$

Since $330^o=360^o+330^o=690^o
$ and $330^o=2\times 360^o+330^o=1050^o$ there are two more solutions:
1. $3(\cos{230^o}+i\sin{230^o})$
2. $3(\cos{350^o}+i\sin{350^o})$

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2 years ago

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marta [7]

Answer:

3781

Step-by-step explanation:

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(2125.00+1440.00+216=3781)

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2 years ago

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Triangle H N K is shown. Angle H N K is 90 degrees. The length of hypotenuse H K is n, the length of H N is 12, and the length o

Mazyrski [523]

Answer:

Step-by-step explanation:

From the question, we are given a triangle HNK with an angle of 90°

The length of hypotenuse H K is n,

the length of HN is 12

the length of N K is 6.

From the above values, obtained in the question, we can see that this is a right angled triangle.

We are asked to find the length of the hypotenuse.

We can use Pythagoras Theorem of solve for this.

c² = a² + b²

where c = HK = n

a = NK = 6

b = HN = 12

c² = 6² + 12²

c² = 36 + 144

c² = 180

c = √180

c = 13.416407865

Approximately to the nearest whole number = 13

Therefore the value of HK = n = 13

We can also use Law of Cosines as given in the question to solve for this.

a² = b² + c² - 2ac × Cos A

where c = HK = n

a = NK = 6

b = HN = 12

Hence

c² = a² + b² - 2ab × Cos C

c = √ (a² + b² - 2ab × Cos C)

Where C = 90

c = √ 6² + 12² - 2 × 6 × 12 × Cos 90

c = 13.42

Approximately to the nearest whole number ≈ 13

Therefore the value of HK = n = 13

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kkurt [141]

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