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kakasveta [241]
3 years ago
5

How many different rays can be formed from five collinear points?

Mathematics
2 answers:
valkas [14]3 years ago
7 0

Answer:

eight

Step-by-step explanation:

spayn [35]3 years ago
4 0

Answer:

I believe that its infinitely

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Help me at this thanks​...please this is an urgency
Ira Lisetskai [31]

Answer:

if simplified is 30

Step-by-step explanation:

600 x 360/7,200

3 0
3 years ago
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The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

5 0
3 years ago
Emma, Brandy, and Damian will cut a rope that is 32.8 feet long into 3 jump ropes. Each of the 3 jump ropes will be the same len
Zolol [24]

Answer:

10.93 is your answer bud

8 0
3 years ago
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What is the domain of the sine and cosine functions?
eimsori [14]

Answer:

domain of the sine and cosine functions is all real numbers

Step-by-step explanation:

range of both the sine and cosine functions is [−1,1]

graph can go left and right to infinity

graph goes up and down between 1 and -1

graph looks like a wave or

like U letters next to each other

UUUU

sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.

mathlibretextsorg

8 0
2 years ago
Solve 2x2 8x - 12 = 0 by completing the square. show step-by-step solution
malfutka [58]
Group x terms
(2x^2+8x)-12=0
undistribute 2
2(x^2+4x)-12=0
take 1/2 of 4 and square it, add negative and positive of it insde (2^2=4)
2(x^2+4x+4-4)-12=0
factor perfect square
2((x+2)^2-4)-12=0
distribute
2(x+2)^2-8-12=0
2(x+2)^2-20=0
add 20 to both sides
2(x+2)^2=20
divide both sides by 2
(x+2)^2=10
sqrt both sides, remember to take positive and negative root
x+2=+/-√10
minus 2 both sides


x=-2+√10 and -2-√10
8 0
3 years ago
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