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3 years ago

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Lake in East Texas was stocked with 3,000 trout in the year 2000. The number of fish in the lake is modeled by f(x) = 3000 + 100

x, where x represents the number of years after 2000.
Which of the following is the most reasonable domain and range for the function?

2 answers:

krok68 [10]3 years ago

8 0

Answer:

Step-by-step explanation:

A

aivan3 [116]3 years ago

5 0

Answer:A

Step-by-step explanation:A

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Solve the system by the addition method <br> 4x=3y-8<br> -16x+4y=32

butalik [34]

4x = 3y - 8
4    4
x = ³/₄y - 2

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3 years ago

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Step-by-step explanation:

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Step-by-step explanation:

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HELP<br> PLEASE. <br> Simplify the expression.<br> x^2+2x-24/x^2-x-42

Sophie [7]

Answer:

$\frac{x - 4}{x - 7}$

Step-by-step explanation:

$\frac{ {x}^{2} + 2x - 24 }{ {x}^{2} - x - 42 } \\ \frac{(x + 6) (x - 4)}{(x - 7)(x + 6)} \:the \: term \: will \: be \: simplified (x + 6) \\ and \: we \: will \: be \: left \: with \: \frac{x - 4}{x - 7}$

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Read 2 more answers

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of <em>A</em> is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives

$dA=49\cdot 0.1+55 \cdot 0.1$

$dA=10.4$

Thus the maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

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3 years ago