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Reika [66]
2 years ago
10

CAN PLEASE SOMEONE GO TO MY PROFILE AND ANSWER SOME QUESTIONS, IF SOMEONE ANSWERS IT I WILL MAKE ANOTHER "QUESTION" FOR 40 POINT

S BUT NO NONSENSE PLEASE THIS WILL HELP ME ADVANCE IN MATH, BUT FOR NOW, I WILL PUT IT AS 10 POINTS (PLEASE ANSWER MY MOST READ QUESTION THANK YOU)
Mathematics
2 answers:
lilavasa [31]2 years ago
6 0

Answer:

ok

Step-by-step explanation:

Igoryamba2 years ago
6 0

Answer:

le ok

Step-by-step explanation:

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g(5)= (-5)^2 + 5(-2) - 4

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What is the area of this figure?<br><br> Enter your answer in the box.<br><br> in²,
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Step-by-step explanation:

(20×16) + (½×8×20)

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Subtract.<br><br> 3x−5x−4−x2+12x−17x2−16
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Step-by-step explanation:

3x - 5x - 4 - x^2 + 12x - 17x^2 - 16

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Solve this system using substitution:
Sholpan [36]

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4 years ago
Read 2 more answers
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
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