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3 years ago

How can i find the angle between two vectors, given in the format '' i+j ''

Mathematics

1 answer:

vivado [14]3 years ago

5 0

You use the formula below...,

Angle theta = cos^-1 [scalar product of the two vectors ÷ multiplication of their moduli]

For example...the angle between

(2i + 3j) and (4i + 5j)

Angle theta = cos^-1 [( 2(4) + 3(5))÷((square root of (2^2 + 3^2))×(square root of (4^2 + 5^2))

Angle theta = cos^-1 [ 23 ÷ ((square root of(13)) × (square root of (41)]

Angle theta = 5°

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Your plan was to be on the road by 9 A.M. but you did not leave the garage until 10 A.M. You then drove with the cruise control

Wittaler [7]

Answer:

The average speed over the time interval from 9 A.M. to noon was of 33.33 mph.

Step-by-step explanation:

The average speed is given by the following equation:

$v = \frac{d}{t}$

From 10A.M. to noon.

You traveled at 50mph during 2 hours. So how long you traveled?

$50 = \frac{d}{2}$

$d = 50*2$

$d = 100$

You traveled 100 miles.

What was your average speed over the time interval from 9 A.M. to noon

From 9 A.M. to 10 A.M, you did not travel.

From 10 P.M. to noon, 100 miles.

So 100 miles during 3 hours.

$v = \frac{d}{t} = \frac{100}{3} = 33.33$

The average speed over the time interval from 9 A.M. to noon was of 33.33 mph.

8 0

3 years ago

What is the standard form of (7 - 51)(2 + 3)?

patriot [66]

AnswerAnd Step-by-step explanation:

7-5x2+3

x= times or multiplied

3 0

3 years ago

Read 2 more answers

consider the followin polynomials equations a=3x2(x-1) b= -3x3+4x2-2x+1 . perform each operation and determine if the result is

Sedaia [141]

Answer:

All three operations lead to polynomials.

See explanations below.

Step-by-step explanation:

Polynomial a = $3x^2(x-1)=3x^3-3x^2$

Polynomial b = $-3x^3+4x^2-2x+1$

Therefore:

a + b = $3x^3-3x^2+(-3x^3+4x^2-2x+1)=\\=3x^3-3x^2-3x^3+4x^2-2x+1=\\=x^2-2x+1$

where we have combined all like terms. This is clearly another polynomial (of grade 2)

a - b (here we need to flip all signs inside the parenthesis when we remove this grouping symbol): $3x^3-3x^2-(-3x^3+4x^2-2x+1)=\\3x^3-3x^2+3x^3-4x^2+2x-1=\\6x^3-7x^2+2x-1$

which is clearly another polynomial (but of grade 3)

a * b : (here we use distributive property to multiply each term of the first polynomial by each term of the second one, and then combine like terms)

$(3x^3-3x^2)*(-3x^3+4x^2-2x+1)=\\-9x^6+12x^5-6x^4+3x^3+9x^5-12x^4+6x^3-3x^2=\\-9x^6+21x^5-18x^4+9x^3-3x^2$ which is indeed another polynomial (this time of grade 6)