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3 years ago

How do I simplify (2k ^2 +5k-6) (3k-1)

1 answer:

8 0

(2k² + 5k - 6)(3k - 1)
(6k³ - 2k² + 15k² - 5k - 18k + 6)
6k² + 13k² - 23k + 6

Use the FOIL method to simplify the problem.
(FOIL stands for first outer, inner, last)

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Let u = (1,2), v = (−3,4), and w = (5,0)

sergij07 [2.7K]

Answer:

w-2u-v

Step-by-step explanation:

Given are three vectors u, v and w.

In R^2 we treat first element as x coordinate and 2nd element as y coordinate.

Thus we mark (1,2) in the I quadrant, (-3,4) in II quadrant and (5,0) on positive x axis 5 units form the origin.

b) $w=au+bv$

We have to find the values of a and b

$(5,0) = a(1,2)+b(-3,4)$ ]

Equate the corresponding terms

$5=a-3b\\0=2a+4b$

Divide II equation by 2 to get

$0=a+2b$

Eliminate a

-5 = 5b: b=-1

a=-2

Hence

w = 2u-v

7 0

3 years ago

The answer to the problem

kotegsom [21]

37.01 for Jayna.
36.91 for Myriam.
36.78 for Riley.

3 0

3 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$