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Annette [7]
3 years ago
12

The graph shows the influence of the temperature T on the maximum sustainable swimming speed S of Coho salmon. (b) Estimate the

values of S '(5) and S '(25).

Mathematics
2 answers:
Alex17521 [72]3 years ago
3 0
From the graph it appears that S′(5) 2 ≈ and S′(25) 2 ≈ − . The important thing is that they do have opposite signs. The first means that at about 5ºC the Coho gains about 2 cm/sec while at 25ºC it loses about 2 cm/sec in maximum sustainable speed.
ivann1987 [24]3 years ago
3 0

Answer:

From the graph of S, we want to have information about S' (the derivate of S)

For S'(5) we need to look at the beginning of the curve, we can see that S(5) is a value around the 15cm/s, and it increases after that point, so S'(5) must be a positive number, this means that the neighborhood of T = 5°C, the speed will increase as the temperature increases.

For S'(25) we can see that the graph decreases after that point, so S'(25) is a negative number, this means that after the 25°C, if we keep increasing the temperature we will have a decrease in the speed.

We can make an estimation about the values:

now, the average rate of change in an interval can be calculated as:

S' = (Y2 - Y1)/(X2 - X1)

S is almost linear between T = 5°C and 10°C and we have that

S(5) = 15cm/s and S(10) = 20cm/s

so we can find the average rate of change in that interval as:

S' = (20cm/s - 15cm/s)/(10°C  - 5°C) = 1(cm/s°C)

So we can estimate that S'(5) is around 1 (cm/s°C)

We can do something similar for S'(25)

we can estimate that S(20) = 25cm/s and S(25) = 20cm/s

so the average rate of change is:

S' = (20cm/s - 25cm/s)/(25°C - 20°C) = -1 (cm/s°C)

So we can expect that S'(25) is around -1 (cm/s°C)

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3 0
4 years ago
Read 2 more answers
Topic: The Quadratic Formula
Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

x= \frac{-b +- \sqrt{b^2-4ac} }{2a}

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² =  x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\=  -4 or 1

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

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