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prohojiy [21]
3 years ago
13

two runners are saving money to attend a marrathon. the first runner has $112 in savings, received a $45 gift from a friend, and

will save $25 each month. The second runner has $50 in savings and will save $60 each month. Which equation can be used to find m, the number of months it will take for both accounts to habe the same accounts to have the same account of money?
Mathematics
2 answers:
8090 [49]3 years ago
6 0

Answer:

112+45+25m = 50+60m

Step-by-step explanation:

Let m be the number of months it will take for both accounts to have the same account of money

The first runner has $112 in savings, received a $45 gift from a friend, and will save $25 each month.

The equation becomes:

112+45+25m

The second runner has $50 in savings and will save $60 each month.

The equation becomes:

50+60m

Now, to find 'm' we will make both the equations equal.

112+45+25m = 50+60m

So, this is the equation that is your answer.

Arturiano [62]3 years ago
5 0
112+45+25m=50+60m
Both sides should be qual to each other once they both hit a certain number of months. I hope this helps!
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Answer:

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

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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

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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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