All categories

3 years ago

9 - 12n 2 + 4n 3 - 7n

Mathematics

1 answer:

Naddika [18.5K]3 years ago

3 0

1) is the correct answer because you just have to arrange them by degree

You might be interested in

The owner of a small store buys coats for $55.00 ecich. Answer parts a and b.

Darya [45]

Answer: 62.5% 162.5%

and here is more words so the program lets me post this

3 0

3 years ago

What is the area of a rectangular playground with a length of 24 feet and a width of 18 feet?

maw [93]

Answer: 432

Step-by-step explanation:

Multiply your length times your width to get the area.

6 0

3 years ago

Read 2 more answers

Please help with #2 asap !

Triss [41]

Answer:

I'm sorry but I can't see question 2

Step-by-step explanation:

4 0

3 years ago

In the figure, all, and both lines are intersected by transversal & Complete the statements to prove that m≤1 = m25.

m_a_m_a [10]

First blank: Substitution or transitive property of equality

Second blank: Subtraction property of equality

7 0

2 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