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3 years ago

Which expression is equivalent to i 233? a)1 b)–1 c)i d)–i

Mathematics

2 answers:

Aneli [31]3 years ago

5 0

We have to find the value of the expression $i^{233}$

We know that the below values.

$i^2=-1\\i^4=1$

Hence, in order to find the value of the given expression, we can first rewrite it in terms of $i^4$

$i^{233}=(i^4)^{58}\cdot i$

Now, we know that $i^4=1$

Hence, we have

$i^{233}=(1)^{58}\cdot i$

$i^{233}=1\cdot i$

$i^{233}=i$

C is the correct option.

Alinara [238K]3 years ago

5 0

Choice c) I is the answer

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a) P(X∩Y) = 0.2

b) $P_1$ = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

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$P_1$ = 0.36 - 0.2 = 0.16

At the same way, the probability $P_2$ that he must stop at the second signal but not at the first one can be calculated as:

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2 years ago

A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5

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Answer:

A) It will get to a temperature of 125°F at 9:19 PM

B) It will get to a temperature of 150°F at 9:16 PM

C) as time passes temperature approaches the initial temperature of 450°F

Step-by-step explanation:

We are given;

Initial temperature; T_i = 450°F

Room temperature; T_r = 70°F

From Newton's law of cooling, temperature after time (t) is given as;

T(t) = T_r + (T_i - T_r)e^(-kt)

Where k is cooling rate and t is time after the initial temperature.

Now, we are told that After 5 minutes, the temperature is 300°F.

Thus;

300 = 70 + (450 - 70)e^(-5k)

300 - 70 = 380e^(-5k)

230/380 = e^(-5k)

e^(-5k) = 0.6053

-5k = In 0.6053

-5k = -0.502

k = 0.502/5

k = 0.1004 /min

A) Thus, at temperature of 125°F, we can find the time from;

125 = 70 + (450 - 70)e^(-0.1004t)

125 - 70 = 380e^(-0.1004t)

55/380 = e^(-0.1004t)

In (55/380) = -0.1004t

-0.1004t = -1.9328

t = 1.9328/0.1018

t ≈ 19 minutes

Thus, it will get to a temperature of 125°F at 9:19 PM

B) Thus, at temperature of 150°F, we can find the time from;

150 = 70 + (450 - 70)e^(-0.1004t)

150 - 70 = 380e^(-0.1004t)

80/380 = e^(-0.1004t)

In (80/380) = -0.1004t

-0.1004t = -1.5581

t = 1.5581/0.1004

t ≈ 16 minutes.

Thus, it will get to a temperature of 150°F at 9:16 PM

C) As time passes which means as it approaches to infinity, it means that e^(-kt) gets to 1.

Thus,we have;

T(t) = T_r + (T_i - T_r)

T_r will cancel out to give;

T(t) = T_i

Thus, as time passes temperature approaches the initial temperature of 450°F

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