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3 years ago

Why is √3 an irrational number?

Mathematics

1 answer:

anastassius [24]3 years ago

5 0

Answer:

Step-by-step explanation:

B. The number is represented by a non terminating, non-repeating decimal.

(Strictly speaking it is irrational because it cannot be expressed as a ratio of two integers. All rational numbers can be expressed as a ratio of two integers.)

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Solve the equation for x. x/3 = -4 A. -12 B.-7 C.7 D.12

UNO [17]

-12 divided by 3 equals -4

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3 years ago

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Maru [420]

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3 years ago

What are the coordinates of the point 3 units from the origin and on the x-axis?

Maksim231197 [3]

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3 years ago

Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k

valkas [14]

Answer:

$\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\$

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k, the vector projection of B unto a is expressed as $proj_ab = \dfrac{b.a}{||a||^2} * a$

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

||a|| = √130

square both sides

||a||² = (√130)

||a||² = 130

$proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\$

Hence the projection of b unto a is expressed as $\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\$

7 0

4 years ago