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WINSTONCH [101]
3 years ago
14

The average annual income, I, in dollars, of a lawyer with an age of x years is modeled with the following function:

Mathematics
1 answer:
Eduardwww [97]3 years ago
8 0
I = -425x^2 + 45500x - 650000
For I to be maximum, dI/dx = 0
dI/dx = -850x + 45500 = 0
x = 45500/850 = 53.53

Therefore, maximum I = -425(53.53)^2 + 45500(53.53) - 650000 = 567,794.11

Therefore, <span>the maximum average annual income, in dollars, a lawyer can earn</span>  is $567,794
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The formula is
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P principle 2500
R interest rate 0.04
T time ?
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T=I÷pr
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T=2years

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The midpoint of PQ is M(2,-3). One endpoint is<br> P(4,1). Find the coordinates of endpoint Q.
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Q (0,-7)

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3 years ago
Evaluate 12∣∣−x+3∣∣−5 when x = 11.
Evgen [1.6K]

Answer:

91

Step-by-step explanation:

12∣∣−x+3∣∣−5

Let x = 11

12∣∣−11+3∣∣−5

Do the absolute value first

12∣∣-8∣∣−5

Taking the absolute value, which means making it positive

12* 8 -5

Multiply and divide

96 -5

91

7 0
3 years ago
HELP i dont get it<br> this is so harddddd
Lady bird [3.3K]
B. the line is going up so it’s going to be positive not negative. the b value is gonna be negative 1 instead of positive because it crosses the y axis closer to -1 than it does 1
4 0
3 years ago
Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

3 0
3 years ago
Read 2 more answers
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