Which function best fits the following points?
A.=-12.84032+0.0225x
O B. y=65.0778-772.9605*
O C. y=-197.0571x2+ 245.6243x + 6.0321
O D. None of the above

Answer: y = 3x+5, 3, (-1,2)

Step-by-step explanation:

y = mx + c where m is the slope, so we must find the slope

Slope = rise/run = (5-2)/(0-(-1)) = 3

So y = 3x + c

To find c, we have to substitute one of the coordinates on the graph into the equation, (-1,2), the point that is marked

2 = 3(-1) + c

c = 5

So the equation of the graph is y = 3x + 5

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

The given number is :

[tex]\qquad❖ \: \sf \: \cfrac{ - 20}{ \sqrt{10} } [/tex]

since 10 isn't a perfect square, root 10 is an irrational number, and therefore the whole number is irrational as well, and we already know that irrational numbers are also part of real numbers.

[tex] \qquad \large \sf {Conclusion} : [/tex]

So the Correct choices are :

The speed of the train is 150 km/hour.

The distance covered by the train in 3 4/5 hours is 570 km.

The speed of a body is calculated using the formula, speed = distance/time.

The distance covered is calculated using the formula, distance = speed*time.

The time taken by a body is calculated using the formula, time = distance/speed.

In the question, we are asked for the speed of a train, when it covers a distance of 225 km in 1 1/2 hours.

Distance = 225 km.

Time = 1 1/2 hours = 1.5 hours.

Speed = Distance/Time = 225/1.5 km/hour = 150 km/hour.

Now, we are asked to calculate the distance covered by the train in 3 4/5 hours.

Speed = 150 km/hour.

Time = 3 4/5 hours = 3.8 hours.

Distance = Speed*Time = 150*3.8 km = 570 km.

Thus, the speed of the train is 150 km/hour.

The distance covered by the train in 3 4/5 hours is 570 km.

The provided question is incorrect. The correct question is:

"A train at a uniform speed covers a distance of 225 km in 1 1/2 hours. Find the speed of the train and the distance covered in 3 4/5 hours.

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If 2+3/4 hours is equal to 1+5/5 equal length shift then the single shift is 1.5 hours long.

Given that Paolo helped in the community garden for 2 +3/4 hours this week which is 1+5/6 times length shift.

We are required to find the length of a single shift.

First we have to find the proper fraction of the given mixed fractions.

Fraction is combination of numerator and denominator.The number that is present above the division sign is known as numerator and the number that is present below the division sign is known as denominator.

2+3/4=(8+3)/4=11/4

1+5/6=(6+5)/6=11/6

let the length of a shift is x hours.So,

11/4=11*x/6

11/4=11/6 x

66/44=x

x=3/2

x=1.5 hours.

Hence if 2+3/4 hours is equal to 1+5/5 equal length shift then the single shift is 1.5 hours long.

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Answer:

YES! it's one-to-one function.

Step-by-step explanation:

One to one function or one to one mapping states that each element of one set, say Set (A) is mapped with a unique element of another set, say, Set (B), where A and B are two different sets. It is also written as 1-1. In terms of function, it is stated as if f(x) = f(y) implies x = y, then f is one to one.

Thus, when we look at this given function

X Y

[tex] - 2 \: \: \: \: \: \: 1 \\ - 5 \: \: \: \: \:- 1 \\ 3 \: \: \: \: \: \: - 5 \\ 1 \: \: \: \: \: \: - 2 \\ 0 \: \: \: \: \: \: 5 \\ - 1 \: \: \: \: \: \: 6 \\ - 6 \: \: \: \: \: \: \: 7 \\ \: \: 7 \: \: \: \: \: \: - 6[/tex]

So, when you look at each value of x is attaches with different value of y and there is no repetition of elements means that one element of x is directly attached to one element of y.

Hope it helps!

If the mean, median, and mode of the five numbers 5, 7, 8, A, and B are equal, the possible values of A+B are 20, where A is 8, and B is 12.

The measures of central tendency include the mode, the median, and the mean.

Statistically, a central tendency is a typical value for a probability distribution.

The mode's value appears most often in a set of data values.  The median is the middle value of a dataset.  The mean represents the average.

To determine the possible values of A and B, we can start by finding the median value, which is the middle value in ascending order.

Median (Middle value) = 8

Mode = 8

Total value = 40 (8 x 5)

A = 8

B = 12.

Mean = 8 (40/5)

Thus, if the mean, median, and mode of the five numbers 5, 7, 8, A, and B are equal, the possible values of A+B are 20, where A is 8, and B is 12.

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For this, let's go through each problem carefully and step-by-step.

According to the question, the rate of change of the temperature of any object that is defined by T, is directly proportional to the difference of T and the temperature of the environment around it, which we'll denote as X.

[tex]\frac{dT}{dt}[/tex][tex]= k (T-X)[/tex]

K is a constant of proportionality here. And the temperature of the surrounding environment is said to be (-18°C). Thus,

[tex]\frac{dT}{dt} = k(T+18)[/tex].

For part A, in order to find the differential equation for T, we need to solve for k. So we separate the variables and then integrate to solve the equation.

[tex]\int\limits{\frac{dT}{T+18} } = \int\limits {k} \, dt[/tex]

[tex]ln(T+18) = kt+c[/tex]

Now thw inital temperature of a pot of chili is 23°C, so at [tex]t = 0, T_0 = 23*C[/tex].

Substituting 23 for T and 0 for t, we have the following:

[tex]ln(23+18) = k(0)+c[/tex]

[tex]ln(41) = c[/tex]

We know the temperature of chili after 2 hours is 7°C, so we know that when [tex]t = 2, T_1 = 7[/tex]

Substituting t for 2, and T for 7, we get:

[tex]ln(7+18) = 2k+ln(41)[/tex]

[tex]ln(25) = 2k + ln(41)[/tex]

Solving for 2k

[tex]2k = ln(25) -ln(41)[/tex]

[tex]2k = ln(\frac{25}{41})[/tex]

[tex]k = \frac{1}{2}ln(\frac{25}{41})[/tex].

Substituting the value of [tex]\frac{dT}{dt} = k (T+18)[/tex], the differential equation obtained is [tex]\frac{dT}{dt} = \frac{1}{2}ln(\frac{25}{41})(T+18)[/tex].

For part B, to find the temperature of the chili after four hours, we first need to solve the above differential equation.

The solution of the differential equation is given by the equation [tex]ln(T+18) = kt+c[/tex]. Substituting the values of k and c, we have:

[tex]ln(T+18) = \frac{1}{2}ln(\frac{25}{41})t+ln(41)[/tex].

Using the above relation, at any time (t), the temperature (T) can be found out in the following.

At [tex]t = 4, T_2 = \phi[/tex]

[tex]ln(T_2+18)=\frac{1}{2}ln(\frac{25}{41})*4+ln(41)[/tex]

[tex]ln(T_2+18)=2ln(\frac{25}{41})+ln(41)[/tex]

[tex]ln(T_2+18)=-0.989 + 3.714[/tex]

[tex]ln(T_2+18)[/tex] ≅ [tex]2.725[/tex]

Solving the natural logarithm,

[tex]T_2+18 = e^{2.725} = 15.256[/tex]

[tex]T_2 =15.256 - 18[/tex]

[tex]T_2 = -2.744[/tex].

So the temperature of the chili after four hours would be -2.744°C approximately.

To find part C in what time the chili would be 10°C, we need to substitute again.

[tex]t = \phi[/tex][tex], T = -10[/tex]

[tex]ln(-10 + 18) = \frac{1}{2}ln(\frac{25}{41})t + ln(41)[/tex]

[tex]ln(8) = \frac{1}{2}ln(\frac{25}{41})t + ln(41)[/tex]

Solving for [tex]\frac{1}{2}ln(\frac{25}{41})t[/tex],

[tex]\frac{1}{2}ln(\frac{25}{41})t = ln(8) - ln(41)[/tex]

[tex]\frac{1}{2}ln(\frac{25}{41})t = ln(\frac{8}{41})[/tex]

[tex]\frac{1}{2}ln(\frac{25}{41})t = -1.634[/tex]

[tex]ln(\frac{25}{41})t[/tex][tex]= -1.634 * 2[/tex]

[tex](-0.494)t=-3.268[/tex]

[tex]t = \frac{-3.268}{-0.494}[/tex]

[tex]t=6.615[/tex] hours, approximately.

Thus, the chili would reach -10°C at around 6.615 hours.

Hope this helped. This took me a long time.

Answer:

Gradient of the line is choice D.9/5

Step-by-step explanation:

Slope between two points:slope=(y₂-y₁)/(x₂-x₁)

slope(m)=

[tex] \frac{5 - ( - 4)}{3 - ( - 2)} \\ refine \: (m )= \frac{9}{5} [/tex]