MATHEMATICS HIGH SCHOOL

The radius r of a circle can be written as a function of the area A with the following equation: What is the domain of this function? Explain why it makes sense in this context.

Answers

Answer 1

Answer:

Hence, the domain of the function is:

[0,∞)

Step-by-step explanation:

We know that area of circle is given by the function:

$A=\pi r^2$

The radius r of a circle can be written as a function of the area A with the following equation:

Now we can represent r in terms of A as:

$r^2=(A)/(\pi)\n\nr=\sqrt{(A)/(\pi)}$

Now as we know that for the square root term to exist:

$\sqrt{(A)/(\pi)}\geq0$

i.e. $A\geq0$

A=0 represents a point circle since it's area is zero.

Hence, the domain of the function is:

[0,∞)

Answer 2

Answer: [0,infinity) because you can not take the radius of a negative mumber.

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What is 43% more than 600? Please show work

Answers

Answer:

143/100*600=858

$600+43\%*600=600+0,43*600=600+258=858\n\n43\%\more\ than\ 600\ is\ 858.$

Leah has 6 1/4 yards of ribbon to make bows. Each bow is made from a piece of ribbon that is 1/3 yard long. What is the maximum number of complete bows Leah can make?

Answers

Leah can make 18 complete bows. She will then have 1/4 yards of fabric left over. This is true because there are 3 thirds to a whole and there are 6 complete yards of ribbon. 3 x 6 = 18. Then we get left with an extra 1/4 yards of fabric because 1/4 is less and a third.

Set G is the set of positive integers divisible by 4 and Set F is the set of perfectsquares. List the first 5 elements of set H, which contains numbers in G that are
also elements of F.

Answers

Greetings from Brasil...

G = {4; 8; 12; 16; 20; 24; 28; 32; 36; 40; 44; 48; 52; 56; 60; 64; 68; 72; 76; 80; 84; 88; 92; 96; 100; 104; ...}

F = {1; 4; 9; 16; 25; 36; 49; 64; 81; 100; ...}

So, according to the statement, it is desired:

G ∩ F - the intersection between the 2 sets, that is, which numbers are present simultaneously in the 2 sets....

Looking at the sets we conclude that

G ∩ F = {4; 16; 36; 64; 100}

OBS: note that in truth G are the multiples of 4

Final answer:

The first five elements of set H, which include positive integers divisible by 4 that are also perfect squares, are 4, 16, 36, 64, and 100.

Explanation:

The two sets mentioned in the problem are Set G, which contains positive integers divisible by 4, and Set F, which contains perfect squares. The intersection of these two sets is Set H. To find the elements of Set H, we look for numbers that are both divisible by 4 and perfect squares. The first five such numbers are 4, 16, 36, 64, and 100. For example, 16 is both a multiple of 4 and a perfect square because it can be expressed as 4*4 and is the square of 4. Similarly, 36 fits both criteria because it can be expressed as 4*9 and is the square of 6. We continue this pattern to identify the first five elements of Set H.

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Determine whether each sequence is arithmetic or geometric. Sequence 1: –10, 20, –40, 80, ... Sequence 2: 15, –5, –25, –45, ...A. Sequence 1 is arithmetic and Sequence 2 is geometric.

B. Both sequences are geometric.

C. Sequence 2 is arithmetic and Sequence 1 is geometric.

D. Both sequences are arithmetic.

sequence 2 may be arithmetic because -20 but I can't find out what sequence 1 is

Answers

Hello,

Answer C

-10,20,-40,80,... $u_(n+1)=(-2)*u_(n)$ . is geometric

15,-5,-25,-45... $u_(n+1)=u_(n)-20$ . is arithmetic

Answer:

C. Sequence 2 is arithmetic and Sequence 1 is geometric.

Step-by-step explanation:

Which of the following represents a polygon that is two times the size of the original polygon?A: (x' , y') = (0.2x,0.2y)
B: (x' , y') = (1/2x, 1/2y) <-- Fractions
C: (x' , y') = (4x,4y)
D: (x' , y') = (2x,2y)

Answers

The expression representing two times the size of the original polygon is (x' , y') = (2x,2y), hence, option (D) is the correct answer.

What is dilation?

A dilation is a transformation that yields a picture that differs in size while maintaining the original image's shape.

The size of the polygon will be doubled if its coordinates are multiplied by 2.

The expression representing the two times the size of the original polygon is (x' , y') = (2x,2y)

Hence, option (D) is the correct answer.

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The answer is D. (2x,2y) because when you two times an ordered pair you multiply it by two.

Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas. He starts by assigning coordinates as given.A right triangle is graphed on a coordinate plane. The horizontal x-axis and y-axis are solid, and the grid is hidden. The vertices are labeled as M, K, and L. The vertex labeled as M lies on begin ordered pair 0 comma 0 end ordered pair. The vertex labeled as K lies on begin ordered pair 0 comma 2 b end ordered pair. The vertex labeled as L lies on begin ordered pair 2a comma 0 end ordered pair. A bisector is drawn from point M to the line KL. The intersection point on line KL is labeled as N.

Enter the answers to complete the coordinate proof.
N is the midpoint of KL¯¯¯¯¯KL¯ . Therefore, the coordinates of N are (a,
).

To find the area of △KNM△KNM , the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM△KNM is
.

To find the area of △MNL△MNL , the length of the base ML is
, and the length of the height is
. So an expression for the area of △MNL△MNL is ab.

Comparing the expressions for the areas shows that the areas of the triangles are equal.

Answers

The coordinates of N is (a,b) using the midpoint formula.
The area for △KNM is (1/2)(a)(2b) = ab
The area of △MNL is ab.
Since the area of △KNM = △MNL and the area of △KML is 2ab, then we have proved that a segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.

1. N is a midpoint of the segment KL, then N has coordinates

$\left((x_K+x_L)/(2),(y_K+y_L)/(2) \right) =\left((0+2a)/(2),(2b+0)/(2) \right) =(a,b).$

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

$A_(KMN)=(1)/(2)\cdot \text{base}\cdot \text{height}=(1)/(2)\cdot 2b\cdot a=ab.$

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

$A_(MNL)=(1)/(2)\cdot \text{base}\cdot \text{height}=(1)/(2)\cdot 2a\cdot b=ab.$

4. Comparing the expressions for the areas you have that the area $A_(KMN)$ is equal to the area $A_(MNL)$ . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.

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