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Nataly [62]
3 years ago
9

If an object moving at 1 mile/hour accelerates 1 mile/hour for every meter it travels, how fast will it be travelling in half of

an hour? (Note: There are 1609.34 meters in a mile)
Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

Speed = 0.0003miles/hr

Step-by-step explanation:

Speed of object = 1 mile/hour

For every meter it travels, its acceleration increases by 1 mile/hour

∆Distance = increase in distance

∆speed = 1 meter

The speed and the distance are constant

Speed = distance/time

Time = distance/speed = 1meter/(1 miles/hr)

Convert meter to miles

1609.34 meters = 1mile

1meter = (1meter × 1mile)/(1609.34 meters)

1 meter = (1/1609.34)miles

Time = distance/speed = (1/1609.34)miles/(1 miles/hr)

Time = (1/1609.34)hr

We are to determine the speed the object would be travelling in half of an hour?

In 1hr, it travels (1/1609.34)miles

In ½hr, it travels = (1/1609.34)(½)hr

Speed = 0.0003miles/hr

It will be travelling 0.0003miles/hr in half of an hour

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In mathematics, the Nth harmonic number is defined to be 1 + 1/2 + 1/3 + 1/4 + ... + 1/N. So, the first harmonic number is 1, th
Andreyy89

Answer:

cl   +   1/(N+1)

Step-by-step explanation:

If we assume that the Nth harmonic number is cl. Then we are assuming that 1+1/2+1/3+1/4+...+1/N=cl

And we know that the (N+1)th harmonic number can be found by doing

1+1/2+1/3+1/4+...+1/N+1/(N+1)

=cl    +   1/(N+1)

The (N+1)th harmonic number is cl   +   1/(N+1) given that the Nth term is cl

Other way to see the answers:

Maybe you want to write it as a single fraction so you have

[cl(N+1)+1]/(N+1)=[cl*N+cl+1]/(N+1)

3 0
3 years ago
Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution wi
Illusion [34]

Answer:

(a) E(X) = 1

(b) \sigma_X =1

(c)  P(X ≤ 1) = 0.6321

(d) P(2 ≤ X ≤ 5) = 0.1286

Step-by-step explanation:

From the given information; Let consider X to be the time between two successive arrivals at the drive-up window of a local bank.

However; If X is regarded as the exponential distribution with λ = 1 which is identical to a standard gamma distribution with ∝ = 1

The objective is to compute the following :

(a) The expected time between two successive arrivals is;

E(X) = \dfrac{1}{\lambda}

E(X) = \dfrac{1}{1}

E(X) = 1

(b)  The standard deviation of the time between successive arrivals is;

\sigma_X = \sqrt{\dfrac{1}{\lambda^2}}

\sigma_X ={\dfrac{1}{\lambda}}

\sigma_X ={\dfrac{1}{1}}

\sigma_X =1

(c)  P(X ≤ 1)

P(X ≤ 1) = 1 - e⁻¹

P(X ≤ 1) = 0.6321

(d) P(2 ≤ X ≤ 5)

P(2 ≤ X ≤ 5) = [1 -  e⁻⁵] - [1 - e⁻²]

P(2 ≤ X ≤ 5) = 0.9933 - 0.8647

P(2 ≤ X ≤ 5) = 0.1286

7 0
3 years ago
WILL GIVE BRAINLIEST! NEED INSTANT HELP! NEED ANSWER FAST!
saveliy_v [14]

Answer:

vi = 16ft/s

Step-by-step explanation:

96 = 16(2)² + 2v

2v = 96 - 64

2v = 32

v = 16

5 0
3 years ago
Which equation represents the graph below?
tamaranim1 [39]
I am fairly certain it is C
3 0
3 years ago
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Write a polynomial function of least degree with integral coefficients that has the given zeros. -5, 3i
Virty [35]

Solving for the polynomial function of least degree with integral coefficients whose zeros are -5, 3i

 

We have:
x = -5

Then x + 5 = 0

Therefore one of the factors of the polynomial function is (x + 5)


Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0

Therefore one of the factors of the polynomial function is (x^2 + 9)


The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)

= x^3 + 9x + 5x^2 = 45

Therefore, x^3 + 5x^2 + 9x – 45 = 0

f(x) = x^3 + 5x^2 + 9x – 45

 

The polynomial function of least degree with integral coefficients that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45

5 0
3 years ago
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