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3 years ago

1) Find the quotient:(2x4 - 3x2 - 1) = (x - 1)

Mathematics

1 answer:

BARSIC [14]3 years ago

4 0

Answer:

The answer is 2!!!

Step-by-step explanation:

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Translate this sentence into an algebraic equation. 8 less than the product of 5 and y is 16

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The answer is 8-5+y=16

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How do you connect the ideas of congruency and rigid motion and how do you prove congruency?

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3 years ago

Write an expression to represent:<br> 6 times the difference of 5 and 2.

Keith_Richards [23]

Answer:

See below

Step-by-step explanation:

The difference of 5 and 2

5-2 <===== now multply it by 6

6 x (5-2)

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2 years ago

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Select equivalent or not equivalent for each pair of expressions

marissa [1.9K]

Answer:

1: equivalent

2: equivalent

3: not equivalent

Step-by-step explanation:

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3 years ago

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p

GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

$t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2$

you replace t1 from the first equation in the second equation:

$(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}$

Then, for t2 = 0 and t2=1/2 you obtain for t1:

$t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2$

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths intersect are:

$r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)$

B) You have the following two trajectories of two independent particles:

$r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)$

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

$t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t$

From the first equation you have:

$t=1+2t\\\\t=-1$

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0

3 years ago

1) Find the quotient:(2x4 - 3x2 - 1) = (x - 1)​

1) Find the quotient:(2x4 - 3x2 - 1) = (x - 1)