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Lady bird [3.3K]
2 years ago
10

The temperature is changing at a constant rate of –0.75ºC per half hour. At 6:45 PM, the temperature was –4.6ºC. At what time wa

s the temperature –2.1ºC? Be sure to include AM or PM in your answer.
Mathematics
1 answer:
Andrews [41]2 years ago
7 0

Answer:

5:05 PM

Step-by-step explanation:

Each 30 minutes, we have a change of -0.75

So in an hour , we will have a change of -0.75 * 2 = -1.5

since -2.1 is greater than -4.6, this indicates that it was a time before 6:45 PM

so, let us find the temperature drop

That would be;

-2.1-(-4.6) = -2.1 + 4.6 = 2.5

So let us find the number of half hours

That would be;

2.5/0.75 = 3 1/3 half hours

3 half hours is 90 minutes

1/3 half hour is 1/3 * 30 = 10 minutes

Total number of minutes is;

90 + 10 = 100 minutes

So let us count 100 minutes before 6:45 PM

120 minutes before would be;

4:45 PM

but we want 100 minutes, so we have to add 20 minutes

That would be;

4:45 PM + 20 minutes

= 4:65 which is 5 : 05 PM

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Arc AB measures 3(16) = 48 degrees.

Arc BC measures 2(16) + 5 = 41 degrees.

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Step-by-step explanation:

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Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0
Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\

Let   the value a so, the value of a_n  and the value of a_(n+1)is:

\to  a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}

\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}

To calculates its series we divide the above value:

\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\

           = \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |

           = \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |

           = \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\

           = \frac{x^2}{2^2(n+1)^2}\longrightarrow 0   for all x

The final value of the converges series for all x.

8 0
3 years ago
Consider triangle ABC with M < C= 65, b = 5 and c = 6. Which option lists an expression that is equivalent to m <B?
tiny-mole [99]
Given:
m∠C = 65°
b = 5
c = 6

Apply the Law of Sines.
sin(B)/b = sin(C)/c
sin(B)/5 = sin(65)/6
sin(B) = (5*sin(65))/6 = 0.7553
B = arcsin(0.7553) = 49.05°

Answer: m∠B = arcsin(0.7553) = 49° (approx)
3 0
3 years ago
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