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kondor19780726 [428]
3 years ago
12

given directed line segment AB find the coordinates of p such that the ratio of ap to pb is 2:1 plot point p

Mathematics
1 answer:
murzikaleks [220]3 years ago
3 0

Answer:

P(\frac{10}{3}, \frac{-2}{3})

Step-by-step explanation:

The question is incomplete; However

A = (-2, -4)

B = (6, 1)

Required

Determine coordinates of P

When line segment is divided in ratios, the following formula calculates the coordinates;

P(x,y) = \{\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\}

In this case;

m:n = 2:1

(x_1,y_1) = (-2, -4)

(x_2,y_2) = (6, 1)

So, the coordinates of P is calculated as thus

P(x,y) = \{\frac{2 * 6 + 1 * (-2)}{2+1}, \frac{2 * 1 + 1 * (-4)}{2+1}\}

P(x,y) = \{\frac{12 -2}{3}, \frac{2 -4}{3}\}

P(x,y) = \{\frac{10}{3}, \frac{-2}{3}\}

Hence, the coordinates of P is

P(\frac{10}{3}, \frac{-2}{3})

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Two planes are driving toward each other with the speeds 250 mph and 420 mph respectively. If the distance between them are 1250
lakkis [162]

Answer:

They will meet in 1.866 hours.

Step-by-step explanation:

This is actually quite a simple question once we think about it in a different way.

Instead of thinking of two different planes that are going at two different speeds, think of it like this: the distance between the two planes is decreasing by 670mph (250mph + 420 mph).

Note: miles per hour is equivalent to miles/hour (<em>miles divided by hours</em> or <em>miles over hours</em>)

670m/h becomes our new rate. The rate the distance between the planes is decreasing.

Now, if the distance between them is 1250 miles, how long until they meet?

If we have miles and multiply it by hours/miles, the miles cancel out and you are left with hours.

We have miles (1250) and miles/hours (670). If we multiply these we get units of miles^2 / hours which is not what we want.

we need to multiply 1250 miles by something with the units of hours/miles.

It turns out that the reciprocal of miles/hours is hours/miles

That means we need to multiply 1250 miles by the reciprocal of 670 miles/hours

Note: the reciprocal of a fraction is the fraction flipped around. 1/2 becomes 2/1, reciprocal of 3/5 is 5/3, etc...

the reciprocal of 670 miles/hours = 1/670 miles/hours

Multiplying 1250 miles by 1/670 miles/hours is the SAME as

dividing the number of miles(1250m) by the rate(670m/h)

1250m / 670m/h = 1250m x 1/670 x h/m = 1.866 h = 1.866 hours (the miles cancel out)


<h2>Note: I explained it so you know why dividing 1250miles by 670mph gives you hours. Once you understand why, all you need to do in the future is divide the distance by the rate and you get time. </h2><h2>Distance / rate = time</h2><h2 />
8 0
3 years ago
Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Spr
NeX [460]

Answer:

20.2057 Units.

Step-by-step explanation:

First, we determine the length of the cable.

Distance between Centerville (8,0) and point (x,0) is given as:  

  • \sqrt{(8-x)}^2=8-x

Distance between point (x,0) and Springfield(0,7) is:

\sqrt{(7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}

Distance between point (x,0) and Shelvyfield(0,-7) is:

\sqrt{(-7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}

Therefore the Length of the Cable L(x)

  • L(x)=(8-x)+2\sqrt{(7)^2+(x)^{2}}

To find the critical point, we set the derivative of L(x)=0

L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}

\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993

Since L^{''}(x)  is positive, the minimum point of L(x) exists.

Next, we find the minimum length by substituting z=3.1305 into L(x)

L(3.1305)=(8-3.1305)+2\sqrt{(7)^2+(3.1305)^{2}}

Minimum Length, L=20.2057

The minimum length of the cable is 20.2057 Units.

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