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3 years ago

PLEASE SOMEONE HELP ME ON THIS

Mathematics

2 answers:

goldenfox [79]3 years ago

7 0

I think the answer for D(n) = 50 ( 3(2) - 2(3) )
Let me know if I'm wrong
Hope it helps ;))

Maksim231197 [3]3 years ago

5 0

Start by figuring out which number is bigger.
f(3) = 150*(2)^3
f(3) = 150* 8
f(3) =1200

g(3) = 100*(3)^3
g(3) = 100*27
g(3) = 2700

Conclusion: g(n) is larger. That's not really the question. The question is how do you enter the data? What they are trying to make you do is stop at the end of the first line and take out the common factor like this
g(n) - f(n) = 100*3^n- 150*2^n The common factor is 50, so it looks like this.
g(n) - f(n) = 50 * (2*3^n - 3*2^n)

So the small box on the left should be 50
The box in brackets should be (2*3^n - 3*2^n)

Where does this course come from? The problem in not math. The problem is reading.

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Solve for x. + 6 = 10 4 1 16 64

Alecsey [184]

Answer: First, turn the mixed fraction 16 1/4 into a improper fraction. It would become 65/4

x+65/4=64

Next, I would multiply the entire equation by 4 to get rid of the fraction

It would become 4x+65=256

Now solve for X

4x=191

x=191/4

x=47.75

Step-by-step explanation:

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3 years ago

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Tell whether x and y show direct variation. x + y = 6

Over [174]

<h3>Answer: Not direct variation</h3>

The reason why is because it can't be written in the form y = kx

If we solved the given equation for y, we get y = -x+6. It has a y intercept of 6, but it should be zero if we wanted a direct variation equation.

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2 years ago

Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1

nevsk [136]

Answer:

a) $u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

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

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) $u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

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

c) $u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each 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.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

$u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

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:

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

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

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

$u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

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:

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

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

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

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

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

$u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

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4 years ago

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Where are the asymptotes of f(x) = tan(4x − π) from x = 0 to x = pi over 2 ?

vladimir1956 [14]

$tan(4x- \pi )= \frac{sin(4x- \pi )}{cos(4x- \pi )}$

The asymptotes are where the graph is undefined. Since: tan(x) =sin(x)/cos(x)
It is where cos(4x-π) = 0

cos(4x-π) = 0 when the inside is -π/2 , π/2 , 3π/2

4x - π = π/2
4x = π/2 + π
4x = 3π/2
x = 3π/8

4x - π = 3π/2
4x = 3π/2 + π
4x = 5π/2
x = 5π/8
This ones outside the interval (5π/8 > π/2) , try -π/2

4x - π = -π/2
4x = -π/2 + π
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x = π/8

Asymptotes are π/8 and 3π/8

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4 years ago

A bicycle tire has a radius of 13 1/4 inches. How far will the bicycle travel in 40 rotations of the tire? Round to the nearest

Sergeu [11.5K]

Answer:

the bicycle will travel 277.5 feet or 3,330.1 inches.

Step-by-step explanation:

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so if you multiply 13 1/4 by 2 you get 26 1.5 if you multiply that by pi than you get the circumference. then multiply this by 40 to get the inches traveled and divide that by 12 to get the number of feet traveled.

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