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3 years ago

Pls help .............

Mathematics

1 answer:

Zolol [24]3 years ago

8 0

The answer is A. k/6 + 8 - 2/k

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Find f(g(x)) f(x)=2x-5 , g(x)=4x+3

hichkok12 [17]

Set the composite results function.

$\boldsymbol{f(g(x)) }$

Evaluate f(g(x)) by substituting the value of g into f.

$\boldsymbol{2(4x+3)-5}$

Simplify each term.

Apply the distributive property.

$\boldsymbol{f(4x+3)=2(4x)+2\cdot3-5 }$

Multiply 4 by 2.

$\boldsymbol{f(4x+3)=8x+2\cdot3-5 }$

Multiply 2 by 3.

$\boldsymbol{f(4x+3)=8x+6-5 }$

Subtract 5 from 6.

$\boldsymbol{f(4x+3)=8x+1 }$

$\red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}$

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1 year ago

Rita ran 5 miles in 48 minutes what was her time per mile

Free_Kalibri [48]

9.60 minutes per mile

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3 years ago

Read 2 more answers

PLEASSSE HELLPP

Orlov [11]

Answer:

C : 0

Step-by-step explanation:

First find f(3). That value is 2.

Then g(f(3)) = 0.

3 0

3 years ago

Find the measure of <BMD (please this is urgent)

Bogdan [553]

Answer:

<BMD = 188°

Step-by-step explanation:

5g + 193

g = 193 - 5

g = 188°

7g + 107

g = 107 - 7

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2 years ago

The lifetime of two light bulbs are modeled as independent and exponential random variables X and Y, with parameters lambda and

Scorpion4ik [409]

Answer:

$\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)$

Step-by-step explanation:

The PDF of X is

$\bf f(x)=\lambda e^{-\lambda x}\;(x\geq0)$

The PDF of Y is

$\bf g(x)=\mu e^{-\mu x}\;(x\geq0)$

The means of X and Y are respectively,

$\bf \displaystyle\frac{1}{\lambda}\;,\displaystyle\frac{1}{\mu}$

so we can see that the larger the parameter, the smaller the mean. Hence the PDF of Z = min(X, Y) is an exponential with the largest parameter of the two.

Therefore, the PDF of Z is

$\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)$

7 0

2 years ago

Pls help .............​

Pls help .............