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2 years ago

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The equation of line is y = x. The midpoint of BC is (a + c, b). Does the midpoint of BC lie on ? Why or why not? no, because b

does not equal a + c no, because (a + c) does not equal b yes, because b = a + c yes, because (a + c) = b

2 answers:

andrew11 [14]2 years ago

6 0

The answer is 2 and 9 because 9 divide by that is 4.5

AleksAgata [21]2 years ago

6 0

Answer:

yes, because (a+c) =b

Step-by-step explanation:

Given that equation of the line is y=x.

B and C are two points lying on this line.

Midpoint of B and C are given as (a+c, b)

From this we see that using midpoint formula

$a+c=\frac{x1+x2}{2 } and\\b=\frac{y1+y2}{2 }$

Since both points lie on y=x, we get

x1=y1 and x2=y2

So x1 = a+c and 2y1 = 2b or y1 =b

Since x1=y1, we have a+c =b

i.e. the coordinates of mid point satisfy the equation y=x

Hence answer is

yes, because (a+c) =b

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2 years ago

Read 2 more answers

Eudora ran from her home to her secret laboratory at an average speed of 12\text{ km/h}12 km/h12, start text, space, k, m, slash

OleMash [197]

Answer:

Eudora spent 0.5 hours from home to library and 1.5 hours from library to school

Step-by-step explanation:

Given

Home to Library:

$S_1 = 12km/h$ -- Average Speed

Library to School

$S_2 = 76km/h$ -- Average Speed

Total

$Distance = 120km$

$Time = 2\ hr$

Required

Determine the time taken from home to library and from library to school

Let the time spent from home to library be x. So, from library to school will be: 2 - x

So, we have:

Home to Library:

$S_1 = 12km/h$ -- Average Speed

$T_1 =x$ -- Time

Library to School

$S_2 = 76km/h$ -- Average Speed

$T_2 =2 - x$ -- Time

Distance is calculated as:

$Distance = Speed * Time$

Home to Library:

$D_1 = S_1 * T_1$

$D_1 = 12 * x$

$D_1 = 12x$

Library to School:

$D_2 = S_2 * T_2$

$D_2 = 76 * (2- x)$

$D_2 = 152- 76x$

The total distance is 120km. So, we have:

$120km = D_1 + D_2$

$120 = 12x + 152 - 76x$

Collect Like Terms

$120 - 152 = 12x - 76x$

$-32 = -64x$

$-64x=-32$

Solve for x

$x = \frac{-32}{-64}$

$x = 0.5$

Recall that:

$T_1 =x$

$T_2 = 2 - x$

So, we have:

$T_1 = 0.5$

$T_2 = 2 -0.5 = 1.5$

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1 year ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

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Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

$-2y\lambda = -2z\lambda$ . If lambda is not zero, then y =z. But, since $-2y\lambda=0$ and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

$\lambda = y^2z^2 = x^2z^2$ which implies $x^2=y^2$

Also, from the first and third equation we have that

$\lambda = y^2z^2 = x^2y^2$ which implies $x^2=z^2$

So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

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2 years ago

The following data represent weights in kilograms of maize harvest from a random sample of 72 experimental plots on St. Vincent,

Scrat [10]

Answer:

See explanation

Step-by-step explanation:

7.8 9.1 9.5 10.0 10.2 10.5 11.1 11.5 11.7 11.8

12.2 12.2 12.5 13.1 13.5 13.7 13.7 <u>14.0 14.4</u> 14.5

14.6 15.2 15.5 16.0 16.0 16.1 16.5 17.2 17.8 18.2

19.0 19.1 19.3 19.8 20.0 20.2 20.3 20.5 20.9 21.1

21.4 21.8 22.0 22.0 22.4 22.5 22.5 22.8 22.8 23.1

23.1 23.2 23.7 <u>23.8 23.8</u> 23.8 23.8 24.0 24.1 24.1

24.5 24.5 24.9 25.1 25.2 25.5 26.1 26.4 26.5 26.7

27.1 29.5

A. The five-number summary is

Minimum = 7.8
Maximum = 29.5
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B. The interquartile range is

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C. See attached diagram

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2 years ago

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Yuki888 [10]

Spinning a roulette wheel 6 times, keeping track of the occurrences of a winning number of "16". Select one:

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The trials are independent. Here in the given experiment each trial is independent of other trial.

From the above consideration, we can clearly say that the given procedure follows binomial distribution.

5 0

2 years ago