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3 years ago

What is 50 - ? = 30

Mathematics

2 answers:

lisov135 [29]3 years ago

6 0

50-20=30

20+30=50

hope this helped you!!!!!

timurjin [86]3 years ago

4 0

Taking 20 away from 50 will result in 30

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Passing through (5, -4) and parallel to the line whose equation is y=-2x+3;<br> point-slope form

Marina CMI [18]

The answer is y=-2x+6

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3 years ago

Explain why 256 + 4 is equivalent to (200 + 4) + (40+4) + (164)

laiz [17]

The expression 256 + 4 is equivalent to (200 + 4) + (40+4) + (16+4) according to distributive and commutative property of addition.

Given the expression 256 + 4. This can be solved using the partial sum expressed as:

256 + 4

= (200 + 40 + 16) + 4

According to the commutative property, A+B = B+A

The arrangement does not affect the result. Hence;

(200 + 40 + 16) + 4 = 4 + (200 + 40 + 16)

Using the distributive law;

4 +(200 + 40 + 16) = (200 + 4) + (40 + 4) + (16 + 4)

Hence the expression 256 + 4 is equivalent to (200 + 4) + (40+4) + (16+4) according to distributive and commutative property of addition.

Learn more on partial sum here: brainly.com/question/6958503

4 0

3 years ago

PLEASE HELP!! DUE SOON!!!

vodomira [7]

Answer:

1. If it is just a line and it has no end points then it means that it is a line if it has endpoints then it's a line segnment.

2. a suplementary angle is equal to 180 degrees and congruent angels are equal angles. (conruent is just a fancy word for equal.)

Step-by-step explanation:

4 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

The quotient of 794.1 divided by 7.61, expressed to two decimal places is?

dangina [55]

794.1/7.61= 104.34954 expressed to two decimal places is 104.35 hope this helps

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3 years ago