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3 years ago

Which of the following is the graph of y = cos(2x)?

Mathematics

2 answers:

irinina [24]3 years ago

5 0

Answer:

Select the one which meets the following characteristics.

Step-by-step explanation:

y = cos(2x)

Characteristics:

Mean line at y = 0

Amplitude = 1

Period = 360/2 = 180° or pi radians

i.e one complete cos cycle is 0 to 180, next is 180 to 360 and so on

Range : from -1 to 1

Colt1911 [192]3 years ago

4 0

Answer:

answer is a on edge2020

Step-by-step explanation:

taking the test, graphed it

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)x= (3x-17) (x+40) (2x-5) solve for x

Eva8 [605]

Answer:

Step-by-step explanation:

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3 years ago

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The ratio of teachers to students on a field trip to New York City is 2 to 9. If there are 243 students on the trip,

Veseljchak [2.6K]

Answer:

54 teachers

Step-by-step explanation:

Step one:

given data

The ratio of teachers to students on a field trip to New York City is 2 to 9.

That is the ratio is 2:9

We are told that there are 243 students

Required

We want to find the number of teachers

Step two:

Applying the part to part method we have

let the number of teachers be x

2/9= x/243

cross multiply we have

9x= 243*2

9x= 486

divide both sides by 9

x= 54

Hence there are 54 teachers

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3 years ago

Describe the translation 7 units to the left, 12 units up using a vector.

Lelu [443]

The magnitude of the vector is √(7^2+12^2)=√(49+144)=√193

The direction is α=arctan(y/x)=arctan(12/-7)≈-59.74 however this is relative to the negative x-axis so we add 180° to get the standard angle...180-59.74=120.26°

So the vector is √193 @ 120.26°

5 0

3 years ago

Factor and solve the equation

murzikaleks [220]

$y=4x^2-64=4(x^2-16)=4(x^2-4^2)=4(x-4)(x+4)\\\\-------------------------------\\\\4(x-4)(x+4)=0\iff x-4=0\ or\ x+4=0\\\\x=4\ or\ x=-4$

8 0

3 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.

5 0

3 years ago

Read 2 more answers