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AlexFokin [52]
2 years ago
14

Multiply (x^2-5x+1)(4x-3x^2)

Mathematics
2 answers:
aleksley [76]2 years ago
6 0

x= 6 is the answer

Step-by-step explanation:

hope it helps

Andreas93 [3]2 years ago
6 0

Answer:

 -x • (x2 - 5x + 1) • (3x - 4)

Step-by-step explanation:

STEP

1: Equation at the end of step 1

 (((x2) -  5x) +  1) • (4x -  3x2)

STEP2:STEP

2:Pulling out like terms

 Pull out like factors :

4x - 3x2  =   -x • (3x - 4)

Trying to factor by splitting the middle term

 Factoring  x2 - 5x + 1

The first term is,  x2  its coefficient is  1 .

The middle term is,  -5x  its coefficient is  -5 .

The last term, "the constant", is  +1

Step-1 : Multiply the coefficient of the first term by the constant   1 • 1 = 1

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -5 .

     -1    +    -1    =    -2

     1    +    1    =    2

Observation : No two such factors can be found !!

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Answer:

a) t = 4

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Step-by-step explanation:

You have the following vector equation for the position of a particle:

r(t)=cos(\pi t)\hat{i}+sin(\pi t)\hat{j}+5t\hat{k}    (1)

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For a height of 20 you have:

5t=20\\\\t=\frac{20}{5}=4

(b) The velocity of the particle is the derivative, in time, of the vector position:

v(t)=\frac{dr(t)}{dt}=-\pi sin(\pi t)\hat{i}+\pi cos(\pi t)\hat{j}+5\hat{k}    (2)

and for t=4 (height = 20):

v(t=4)=-\pi sin(\pi (4))\hat{i}+\pi cos(\pi (4))\hat{j}+5\hat{k}\\\\v(t=4)=-0\hat{i}+\pi\hat{j}+5\hat{k}

(c) The vector parametric equation of the tangent line is given by:

r_t(t)=r_o+vt      (3)

ro: position of the particle for t=4

r_o=cos(\pi (4))\hat{i}+sin(\pi (4))\hat{j}+20\hat{k}\\\\r_o=\hat{i}+0\hat{j}+20\hat{k}

Then you replace ro and v in the equation (3):

r_t=(1\hat{i}+20\hat{k})+(\pi \hat{j}+5\hat{k})t\\\\r_t=1\hat{i}+\pi t \hat{j}+(20+5t)\hat{k}

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77julia77 [94]
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