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2 years ago

The point (1, 4) lies on a circle that is centered at (1, 1). Which statements are correct? Check all that apply.

Mathematics

2 answers:

Semenov [28]2 years ago

6 0

the circle's radius is 3 units

the point (- 2, 1 ) lies on the circle

the equation of a circle in standard form is

(x - a)² + (y - b)² = r²

where (a, b) are the coordinates of the centre and r is the radius

the radius is the distance from the centre to the point (1, 4 ) on the circle

using (1, 4) and (1,1) in the distance formula, then

r = √(1 - 1 )² + (1 - 4)² = √(0 + 9) =√9 = 3 ⇒ r² = 9

(x - 1 )² + (y - 1 )² = 9 ← equation of circle

substitute the given points into the equation and if equation is true then they lie on the circle

(- 2, 1 ) : (- 2 - 1 )² + (1 - 1 )² = 9 + 0 = 9 ← true

Hence (- 2, 1 ) lies on the circle

(3, 3 ) : (3 - 1 )² + (3 - 1 )² = 4 + 4 = 8 ≠ 9

(3, 3 ) does not lie on the circle

Monica [59]2 years ago

4 0

Answer: b and d

Step-by-step explanation:

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Zac and Lynn are each traveling on a trip. So far, Zac has traveled 123.75 miles in 2.25 hours. Lynn leaves half an hour after Z

Savatey [412]

Complete question is;

Zac and Lynn are each traveling on a trip. So far, Zac has traveled 123.75 miles in 2.25 hours. Lynn leaves half an hour after Zac. So far

she has traveled 105 miles in 1.75 hours. Assume Zac and Lynn travel at constant rates.

Let a represent the number of hours that have elapsed since Zac started traveling. Let y represent the number of miles traveled. Write a system of linear equations that represents the distance each of them has traveled since Zac left on his trip.

Assume Zac and Lynn continue to travel at the same constant rates and make no stops.

Determine the solution of the system of linear equations.

Answer:

Zac: y = 55a

Lynn: y = 60(a - ½)

6 hours after Zac started traveling, both Zac and Lynn would have covered 330 miles each.

Step-by-step explanation:

Zac has traveled 123.75 miles in 2.25 hours. Since he travels at constant speed, we can say;

zac's speed = 123.75/2.25 = 55 mi/hr

Similarly, Lynn traveled 105 miles in 1.75 hours. Thus, since she travels at a constant speed;

Lynn's speed = 105/1.75 = 60 mi/hr

Now, we are told that a represents the number of hours that have elapsed since Zac started traveling and y represents the number of miles traveled.

Thus;

a hours after Zac started travelling, his distance covered will be;

Zac: y = 55a

Now,for Lynn, since she started ½ an hour after Zac, it means a hours after Zac started, she had traveled (a - ½) hours.

Thus, Lynn's distance traveled after Zac started = 60(a - ½)

Lynn: y = 60(a - ½)

The solution will be when they have travelled equal distances a hours after Zac started. Thus;

55a = 60(a - ½)

55a = 60a - 30

60a - 55a = 30

5a = 30

a = 30/5

a = 6 hours

Putting 6 for a in y = 55a, we have;

y = 55 × 6

y = 330 miles

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2 years ago

Matthew has three brothers: Alex, Mark, and Luke. The sum of their three ages is 12. List all the different combinations of ages

arsen [322]

Answer:

Step-by-step explanation:

Let Alex's age = x

Mark's age = y

Luke's age = a

x + y + z = 12

The possible combinations for their age are

1. 6, 3, 3

2. 3, 3, 6

3. 4,4, 4

4, 2, 2, 8

5. 8, 2, 2

6. 2, 8, 2

7. 9, 2, 1

Etc the age can keep going as long as you can find any three numbers and add them to give you 12

3 0

2 years ago

The number of species n found on islands typically increases with the area of the island A. Suppose that this relationship is su

Crazy boy [7]

Answer:

$n(A) = n_1A^k$

Step-by-step explanation:

Taking into account that the growth rate of the number of species on the island is proportional to the density of species (number of species between area of the island), a model based on a differential equation is proposed:

$\frac{dn}{dA} = k\frac{n}{A}$

This differential equation can be solved by the method of separable variables like this:

$\frac{dn}{n} = k\frac{dA}{A}$ with what you get:

$\int\ {\frac{dn}{n}}\ = k\int\ {\frac{dA}{A}}$

$ln|n| = kln|A|+C$ . Taking exponentials on both sides of the equation:

$e^{ln|n|} = e^{ln|A|^{k}+C}$

$n(A) = e^{C}A^{k}$

how do you have to $n (1) = n_1$ , then

$n(A) = n_1A^k$

8 0

2 years ago

ΔABC underwent a sequence of rigid transformations to give ΔA′B′C′. Which transformations might have taken place?

Papessa [141]

Answer: The correct option is second, a rotation $90^{\circ}$ clockwise about the origin followed by a reflection across the x-axis.

Explanation:

From the given figure it is noticed that the vertices of ΔABC are A(-6,4), B(-4,6), C(-2,2) and vertices of ΔA'B'C' are A'(4,-6), B(6,-4), C(2,-2).

It means if the point is P(x,y) then after transformation it will be P'(y,x).

If a point P(x,y) reflection across the y-axis followed by a reflection across the x-axis, then the image of point after transformation will be P'(-x,-y), therefore it is not the correct option.

If a shape is rotated $90^{\circ}$ clockwise about the origin then the point P(x,y) will be P'(y,-x) and after that reflect across the x-axis, so the point after transformation will be P'(y,x), therefore it is the correct option.

If a shape is rotated $270^{\circ}$ clockwise about the origin then the point P(x,y) will be P'(-y,x) and after that reflect across the x-axis, so the point after transformation will be P'(-y,-x), therefore it is not the correct option.

If a point P(x,y) reflection across the x-axis followed by a reflection across the y-axis, then the image of point after transformation will be P'(-x,-y), therefore it is not the correct option.

Hence, the correct option is second, a rotation $90^{\circ}$ clockwise about the origin followed by a reflection across the x-axis.

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2 years ago

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A concert promoter sells tickets and has a marginal-profit function given below, where Upper P prime (x )is in dollars per tic

AlexFokin [52]

Answer:

Step-by-step explanation:

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2 years ago