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2 years ago

A segment with endpoints A (4, 2) and C (1, 5) is partitioned by a point B such that AB and BC form a 1:3 ratio. Find B.

Mathematics

1 answer:

Furkat [3]2 years ago

4 0

The coordinates of point B that divides the line in the ratio 1:3 is: ( C )(3.25, 2.75)

<h3>What is a line segment?</h3>

A line segment is formed by joining two points by a straight line.

Analysis:

The coordinates of point B(x,y) are gotten by the formula

x = $\frac{px2 + qx1}{p+q}$

y = $\frac{py2 + qy1}{p + q}$

where, x1 and y1 are coordinates of A and x2 and y2 are coordinates of B

are the values of the ratio p:q.

x1 = 4, x2 = 1 y1 = 5, y2 = 2, p = 1, q = 3

x = $\frac{(1 x 1) + (3 x 4)}{1 + 3}$ = 3.25

y = $\frac{(1 x 5) + ( 3 x 2)}{1+3}$ = 2.75

coordinates of B are ( 3.25, 2.75)

In conclusion, the coordinates of point B are (3.25, 2.75)

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20 points Return to questionItem 4Item 4 20 points Police records in the town of Saratoga show that 13 percent of the drivers st

Sladkaya [172]

Answer:

a) 0.1423

b) 0.2977

c) 0.56

Step-by-step explanation:

For each driver stopped for speeding, there are only two possible outcomes. Either they have invalid licenses, or they do not. The probability of a driver having an invalid license is independent from other drivers. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

13 percent of the drivers stopped for speeding have invalid licenses.

This means that $p = 0.13$

14 drivers are stopped

This means that $n = 14$

(a) None will have an invalid license.

This is $P(X = 0)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.13)^{0}.(0.87)^{14} = 0.1423$

(b) Exactly one will have an invalid license.

This is $P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{14,1}.(0.13)^{1}.(0.87)^{13} = 0.2977$

(c) At least 2 will have invalid licenses.

Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1423 + 0.2977 = 0.44$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.44 = 0.56$

8 0

3 years ago

(csc x - 1) (csc x + 1)

victus00 [196]

<h2>Explanation:</h2>

Here we have the following expression:

$(csc x - 1) (csc x + 1)$

So we need to extend this. Remember that the difference of squares tells us:

$(a-b)(a+b)=a^2-b^2$

So here:

$a=cscx \\ \\ b=1$

Thus:

$(csc x - 1) (csc x + 1)=csc^2x-1 \\ \\ \\ But: \\ \\ cscx=\frac{1}{sinx} \\ \\ \\ Then: \\ \\ csc^2x-1=\frac{1}{sin^2x}-1=\frac{1-sin^2x}{sin^2x}=\frac{cos^2x}{sin^2x}=cot^2x \\ \\ \\ Finally: \\ \\ \boxed{(csc x - 1) (csc x + 1)=cot^2x}$

4 0

3 years ago

Last month, Katie started training for the swim team. She swam 20 laps every day for 18 training days. This month, Katie wants t

nasty-shy [4]

She swim 20 laps every day for 18 training days

so total laps done last month = 20*18 = 360 laps

She will swim 3 more laps per training day and add more training days to her monthly schedule.

20 laps + 3 laps = 23 laps

(a)If Katie adds 5 training days to her schedule this month, will she swim 215 more laps?

18 days + 5 days more = 23 days

23 laps every day for 23 training days

so total laps = 23*23 = 529 laps. That is 169 more laps than 360 laps done last month. Hence she cannot swim 215 more laps

If Katie adds 10 more training days to her schedule this month, will she swim 215 more laps?

18 days + 10 days more = 28 days

23 laps every day for 28 training days

so total laps = 23*28 =644 laps. That is 284 more laps than 360 laps done last month. Hence she can swim 215 more laps

(c)If Katie adds 6 more training days to her schedule this month, will she swim 215 more laps?

18 days + 6 days more = 24 days

23 laps every day for 24 training days

so total laps = 23*24 =552 laps. That is 192 more laps than 360 laps done last month. Hence she cannot swim 215 more laps

Build an equation for finding the number of additional training days she needs to swim 215 more laps

Let x be the number of days more

So training days = 18 + x

215 more laps , so total laps = 360 + 215 = 575

3 more laps per training day so 20 + 3 = 23 laps per day

laps * training days = total laps

so equation becomes

23 * (18+x) = 575

Now solve for x

414 + 23x = 575

Subtract 414 from both sides

23x = 161

x= 7

so she needs 7 training days to swim 215 more laps

3 0

3 years ago

What is the equation of the line that is parallel to the line whose equation is y=-4/3x+7/3 and also passes through the point (-

Artemon [7]

Answer: please help me with this, I will do whatever you want but please help me please!

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The length of a rectangular piece of land is 60 yard more than two times its width. The perimeter is 540 yard. Find its dimensio

babunello [35]

Answer:

l=200 yard

w=70 yard

Step-by-step explanation:

let width =w

length l=2w+60

then perimeter P=2(l+w)=2(2w+60+w)=540

3w+60=270

3 w=270-60=210

w=70

l=2*70+60=200

4 0

3 years ago