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Molodets [167]
3 years ago
6

New York City has a population of 8.55 million and 300 square mile. Manhattan has 1.64 million and 23 square miles. How much gre

ater is the population of Manhattan than New York City?
Mathematics
1 answer:
harkovskaia [24]3 years ago
4 0

Given :

New York City has a population of 8.55 million and 300 square mile area .

Manhattan has 1.64 million and 23 square miles area .

To Find :

How much greater is the population density of Manhattan than New York City.

Solution :

Population density of New York City :

D_{NY}=\dfrac{8.55}{300}=0.029 \ million/mile^2\\\\D_{M}=\dfrac{1.64}{23}=0.071 \ million/mile^2

Now ,

D_{M}-D_{NY}\\\\(0.071-0.029)=0.042\ million/miles^2

Therefore , population density of Manhattan is 0.042\ million/miles^2 greater than New York City .

Hence , this is the required solution .

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Use the power property to rewrite log3x9.
Kitty [74]

Answer:

\log _3\left(x^9\right)=9\log _3\left(x\right)

Step-by-step explanation:

Given the expression

\log _3\:x^9

now rewriting the expression

\log _3\:x^9

\mathrm{Apply\:log\:rule\:}\log _a\left(x^b\right)=b\cdot \log _a\left(x\right),\:\quad \mathrm{\:assuming\:}x\:\ge \:0

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Therefore,

\log _3\left(x^9\right)=9\log _3\left(x\right)

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3 years ago
Read 2 more answers
find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B
SashulF [63]

Answer:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3

|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{ab}{|a| |b|}

And the angle is given by:

\theta = cos^{-1} (\frac{ab}{|a| |b|})

If we replace we got:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

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Answer:

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