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azamat
3 years ago
5

Describe the symmetry of these functions

Mathematics
2 answers:
Zanzabum3 years ago
6 0

Answer:

THe first one has line symmetry only

The second one has rotational symmetry only

Step-by-step explanation:

The line symmetry only has the characteristic of having two equal figures that appear if you put a divisive line in the middle and both parts of the function mirror themselves with the other, rotational symmetry means that if you rotate the function 180º you get the same function with the same shape and measures.

Oduvanchick [21]3 years ago
5 0
If it was line symmetry it would need to repeat the same shape twice, which the first follows but the second doesn’t.
if it was rotational, you would need to be able to take the shape in the top right and rotate it counterclockwise or clockwise to get the shape that locks in place. that doesn’t follow that.
both line and rotational symmetry is incorrect because the first example would need to lock up inside the right side of that example.

the answer is C
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2 years ago
A graphing calculator is recommended. A crystal growth furnace is used in research to determine how best to manufacture crystals
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Answer:

a) 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.

b) 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.

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

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

\delta_2 =|33.449-33.336|=0.113

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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4 years ago
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Answer:

3 hours

Step-by-step explanation:

43-16=27

27/9=3

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3 years ago
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