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malfutka [58]
3 years ago
11

Prove that (a-b) x (a+b)=2a x b is true. (a and b are vectors)

Mathematics
2 answers:
Anon25 [30]3 years ago
6 0
(a-b) X (a+b)
=  aXa - bXa +aXb -bXb (distributing)

Now, cross product of a vector with itself = 0
so, aXa = 0, bXb = 0

Also, aXb = - bXa

so,
(a-b) X (a+b) = 0 + aXb + aXb + 0
= 2aXb

hence, proved :)

RoseWind [281]3 years ago
5 0
( a - b) x ( a + b ) = 2 a + b ( missing arrows above the letters- shows that a and b are vectors)
Left side:
(a x a) + (a x b) - (b x a) + (b x b)= ( a x b ) - ( b x a ) = ( a x b ) +( a x b )=
2 a x b
It is proved to be true.
Explanation: Vector product of the same vectors is 0 (a x a and b x b ). Also, vector product is anticommutative: a x b = - ( b x a ).
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Zielflug [23.3K]
Y is increased by 6 because if you imagine x being 3 then y would = 16 but if you increase x by 3 making 6 then y would = 22 and 16 increased by 6 is 22
8 0
3 years ago
Someone plzzzz help me):
sergiy2304 [10]

C. x³-4x²-16x+24.

In order to solve this problem we have to use the product of the polynomials where  each monomial of the first polynomial is multiplied by all the monomials that form the second polynomial. Afterwards, the similar monomials are added or subtracted.

Multiply the polynomials (x-6)(x²+2x-4)

Multiply eac monomial of the first polynomial by all the monimials of the second polynomial:

(x)(x²)+x(2x)-(x)(4) - (6)(x²) - (6)(2x) - (6)(-4)

x³+2x²-4x -6x²-12x+24

Ordering the similar monomials:

x³+(2x²-6x²)+(-4x - 12x)+24

Getting as result:

x³-4x²-16x+24

3 0
4 years ago
8. How many ways can a committee of 5 students be chosen from a student council of 30 students? Is the order in which the member
tankabanditka [31]

Answer: A committee of 5 students can be chosen from a student council of 30 students in 142506 ways.

No , the order in which the members of the committee are chosen is not  important.

Step-by-step explanation:

Given : The total number of students in the council = 30

The number of students needed to be chosen = 5

The order in which the members of the committee are chosen does not matter.

So we Combinations (If order matters then we use permutations.)

The number of combinations of to select r things of n things = ^nC_r=\dfrac{n!}{r!(n-r)!}

So the number of ways a committee of 5 students can be chosen from a student council of 30 students=^{30}C_5=\dfrac{30!}{5!(30-5)!}

=\dfrac{30\times29\times28\times27\times26\times25!}{(120)\times25!}=142506

Therefore , a committee of 5 students can be chosen from a student council of 30 students in 142506 ways.

4 0
3 years ago
Yo someone deleted my other question, so i will ask a question that 'needs' to be answered. How is your day going. And what is 2
Simora [160]

Answer:

4

Step-by-step explanation:

terrible, but it could be worse

7 0
3 years ago
Read 2 more answers
Find the value for b. 3x^2– X-10 =0<br>​
Greeley [361]

Answer:

3x^2 -(6-5)x-10=0

3x^2 -6x+5x-10=0

3x(x-2)+5(x-2)=0

(x-2) (3x+5) = 0

either,

x -2 =0

x=2

then,

3x +5=0

x= -5/3

3 0
3 years ago
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