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4 years ago

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Mathematics

1 answer:

zhenek [66]4 years ago

4 0

Answer:

1080 = X x Y × Z This is my final answer.

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(a) A linear system can have exactly two solutions. (b) Two systems of linear equations are equivalent when they have the same s

DedPeter [7]

Answer:c) A system of three linear equations in two variables is always inconsistent.

Step-by-step explanation:because it was

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3 years ago

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Alexxandr [17]

Answer: What is the question

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3 years ago

Not a examni promise plz awnser i jeed help

Nostrana [21]

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3 years ago

Read 2 more answers

If you are one of the first 100 people to join a new health club, you are charged a joining fee of $49. Otherwise, you are charg

Drupady [299]

Answer:

a) Total cost = 49 + 38.75m

b) Total cost = 149 + 38.75m

c) The graphs of the lines are parallel and both has slope of - 2.5

d)Difference in total cost = $132.5

Step-by-step explanation:

Total cost , TC = Initial membership fee + monthly charges

a) TC = 49 + 38.75m

b)TC = 149 + 38.75m

d) 149 + 38.75(6months) = 381.5

49 + 38.75(12months) = 514

Difference in total cost = 514 -381.5 = $132.5

7 0

3 years ago

For what values of b are the vectors [−9, b, 5] and [b, b2, b] orthogonal? (Enter your answers as a comma-separated list. If an

aivan3 [116]

Answer:

The values of b are [-2,0,2] for which the vectors are orthogonal.

Step-by-step explanation:

Given:

Vector [-9,b,5] and Vector [b,b^2,b].

According to the question:

Both the above vectors are orthogonal we have to find the value of b.

Orthogonal vectors:

Are the vectors which, are at right angles to each other.
Meaning they are perpendicular to each other.
And their,Dot product is equivalent to zero.

So we have to calculate by dot product and make it equal to zero and then have to solve the b values.

Dot Product:

For vectors a and b where $a=(a_1,a_2,a_3)$ and $b=(b_1,b_2,b_3)$ the dot product is $(a.b)=(a_1b_1+a_2b_2+a_3b_3)$ .

Solving.

⇒ $(-9,b,5).(b,b^2,b)$

⇒ $(-9b+b^3+5b) =0$

⇒ $(b^3-4b)=0$

⇒ $b(b^2-4)=0$ <em>...taking b as common.</em>

⇒ $b(b^2-2^2)=0$

⇒ $b(b-2)(b+2)=0$ <em>...using algebraic identity,</em> $a^2-b^2=(a-b)(a+b)$ .

⇒ Equating with zero individually.

⇒ $b=-2,0,2$

So the values of b are [-2,0,2] for which the given vectors are orthogonal.

8 0

3 years ago