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coldgirl [10]
3 years ago
8

What are two numbers I can use to multiply to 4, but also i can add those numbers to equal 2? Any integers are okay.

Mathematics
1 answer:
LekaFEV [45]3 years ago
8 0

No 'real' numbers can do that job.
The two numbers are complex.

They are ...

       <em>1 + i √3</em>
and
       <em>1 - i √3</em> .


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Step-by-step explanation:

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a) w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896

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b) 202= 0.1w^2 +2.156 w +20

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c)For this case \epsilon = \pm 1 since that's the tolerance 1C

d) \delta_1 =|33.222-33.336|=0.116

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So then we select the smalles value on this case \delta =0.113

e) For this case if we assume a tolerance of \epsilon=\pm 1C for the temperature and a tolerance for the power input \delta =0.113 we see that:

lim_{x \to a} f(x) =L

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

Step-by-step explanation:

For this case we have the following function

T(w)= 0.1 w^2 +2.156 w +20

Where T represent the temperature in Celsius and w the power input in watts.

Part a

For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

203= 0.1w^2 +2.156 w +20

And we can rewrite the expression like this:

0.1w^2 +2.156 w-183=

And we can solve this using the quadratic formula given by:

w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}

Where a =0.1, b =2.156 and c=-183. If we replace we got:

w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336

And since the power can't be negative then the solution would be w = 33.34 watts.

Part b

For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:

202= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-182=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222

204= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-184=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449

So then the range of voltage would be between 33.22 W and 33.45 W.

Part c

For this case \epsilon = \pm 1 since that's the tolerance 1C

Part d

For this case we can do this:

\delta_1 =|33.222-33.336|=0.116

\delta_2 =|33.449-33.336|=0.113

So then we select the smallest value on this case \delta =0.113

Part e

For this case if we assume a tolerance of \epsilon=\pm 1C for the temperature and a tolerance for the power input \delta =0.113 we see that:

lim_{x \to a} f(x) =L

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

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