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3 years ago

BRAINLIEST:

Mathematics

2 answers:

Anettt [7]3 years ago

7 0

Answer:

i think its b tell me if im wrong

Step-by-step explanation:

cricket20 [7]3 years ago

6 0

answer:

c. 2 1/6x + 2 1/6y

step-by-step explanation:

3x−1/2y+2 2/3y−5/6x

solve this to see what you get

3x−1/2y+2 2/3y−5/6x

combine like terms
this will make it easier to solve (its a part of solving this)

3x−1/2y+2 2/3y−5/6x

lets look at the like terms here

−1/2y and + 2 2/3y

3x and - 5/6x

now combine them

−1/2y + 2 2/3y = 13/6y

3x - 5/6x = 13/6x

now look at the original equation and put them together

= 13/6x + 13/6y

= 2 1/6x + 2 1/6y (simplified)

now that we have that simply look at the choices and see which one matches what we got

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In the solution, 3b + 9c = 3(b + 3). What should be done to make it CORRECT?

stiks02 [169]

Answer:

The second term of the binomial factor should be 3c instead of 3.

Step-by-step explanation:

Distributive property:

a*b + a*c = a*(b +c)

3b + 9c = 3*b + 3* 3 *c = 3*(b + 3c)

7 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.

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3 years ago

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Working conditions in the United States in the early 1900's can best be described as?

Dmitry_Shevchenko [17]

Working conditions in the United States in the early 1900’s can best be described as atrocious.

6 0

3 years ago

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Attempt 1 of 1

devlian [24]

Answer:

Yee yee delete this i just wanted points

Step-by-step explanation:

8 0

3 years ago

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Jack and Jill were buying water jugs jack bought 2 medium sized jugs and 3 large sized jugs for 150 dollars and Jill bought 4 me

Mademuasel [1]

Answer: the cost of one medium sized jug is $60

the cost of one large sized jug is $10

Step-by-step explanation:

Let x represent the cost of one medium sized jug.

Let y represent the cost of one large sized jug.

Jack bought 2 medium sized jugs and 3 large sized jugs for 150 dollars. This means that

2x + 3y = 150 - - - - - - - - - - - 1

Jill bought 4 medium jugs and 12 large jugs for 360 dollars. This means that

4x + 12y = 360 - - - - - - - - - - - 2

Multiplying equation 1 by 4 and equation 2 by 2, it becomes

8x + 12y = 600

8x + 24y = 720

Subtracting, it becomes

- 12y = - 120

y = - 120/ - 12

y = 10

Substituting y = 10 into equation 1, it becomes

2x + 3 × 10 = 150

2x + 30 = 150

2x = 150 - 30 = 120

x = 120/2 = 60

3 0

3 years ago