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3 years ago

two chords in the same circle are congruent if and only if the associated central angles are supplementary. true or false

Mathematics

2 answers:

garri49 [273]3 years ago

5 0

Answer: false :)

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DochEvi [55]3 years ago

4 0

Answer:

True

Step-by-step explanation:

just answered on apec

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A group surveyed a mix of people below and above 40 years old about whether or not they visit a dentist once a year. This table

Rama09 [41]

The frequency table that shows the relative frequency of people older than 40 and don't visit the dentist is:

D. Visit Dentist Yearly Don’t Visit Dentist Yearly

Below 40 0.27 0.73

Above 40 0.57 0.43

<h3>What is the relative frequency required?</h3>

The relative frequency of those older than 40 and don't go annually to a dentist can be found as:

= Number of over 40 who don't go to dentist / Total of those over 40 surveyed

Solving gives:

= 13 / (17 + 13)

= 13 / 30

= 0.43

Find out more on relative frequency at brainly.com/question/16832475.

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4 0

1 year 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:

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