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omeli [17]
3 years ago
10

Point A is located at (0, 4) and point B is located at (−2, −3). Find the x value for the point that is 1 over 4 the distance fr

om point A to point B. −1.5 −2 −0.5 −1
Mathematics
2 answers:
Vesna [10]3 years ago
7 0

Answer: -0.5

<u>Step-by-step explanation:</u>

the x-value for A is 0.  

the x-value for B is -2.  

The total distance from A to B is -2.

\frac{1}{4}(-2) = -0.5

  A + -0.5

= 0 - 0.5

=  -0.5

ivolga24 [154]3 years ago
4 0

<u>Answer:</u>

-0.5

<u>Step-by-step explanation:</u>

We are given two points:

A (0, 4) and B (−2, −3)

and we are to find the x-coordinate of the point which is 1/4 the distance from the point A to point B.

Since we only have to find the x coordinate, we will measure the distance of x between point A to B i.e. -2.

So, \frac{1}{4} × -2 = -0.5

Therefore, the x-value for the point that is 1/4 of the distance between point A and B is -0.5.

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Ilia_Sergeevich [38]

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(1 point) Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference betw
slavikrds [6]

Answer:

a. \frac{dT}{dt}=k(T-Tm); T(0)=190

b. C_{0}=122

c. k=-0.00259

d. t=153.39838\\ minutos

Step-by-step explanation:

a. Newton's law of cooling states that the speed with which a body is cooled is proportional to the difference between its temperature and that of the medium in which it is found. Then, the initial value problem is given by:

Tm=68

\frac{dT}{dt}=k(T-Tm); T(0)=190

b. The differential equation obtained is a differential equation of separable variables:

\frac{dT}{T-Tm}=kdt\\\\\int {\frac{dT}{T-Tm}}=\int{kdt}\\\\Ln|T-Tm|=kt+C\\\\T(t)=C_{0}e^{kt}+Tm=C_{0}e^{kt}+68\\\\T(0)=C_{0}e^{k(0)}+68=190\\\\C_{0}=122

c. After 33 minutes of serving the coffee has cooled to 180°:

T(33)=122e^{33k}+68=180\\\\e^{33k}=\frac{112}{122}\\\\33k=Ln(\frac{112}{122})\\\\k=-0.00259

d.

150=122e^{-0.00259t}+68\\\\Ln(\frac{150-68}{122})=-0.00259t\\\\t=153.39838\\\\

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