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sattari [20]
3 years ago
10

Triangle ABC has vertices A(1,4), B(???2,4) , and C(???1,???1) . A dilation with a scale factor of 0.25 and center at the origin

is applied to the triangle. What are the coordinates of C' in the dilated image? Enter your answer in the boxes. PLEASE BE RIGHT
Mathematics
1 answer:
saul85 [17]3 years ago
6 0
Hello,

If you are starting with C as point (1, 1) all you have to do is multiply the coordinates by the scale factor. In this case, the scale factor is 0.25.

So we have:  0.25 x 1 = 0.25

Our new point C' will be (0.25, 0.25)

I hope this helps,
MrEQ 
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aivan3 [116]

Answer:

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Step-by-step explanation:

180=2x+6x+20

180=8x+20

160=8x

20=x

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2 Question help
Travka [436]

Step-by-step explanation:

60 divided by 7 = approx. 8-9 min for each task.

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3 years ago
Read 2 more answers
A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.
Fed [463]

Answer:

A) a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

B) a_{0} = a_{1} = 0

C)   for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a  pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of  consecutive Os therefore there will be : 2^{n-2} strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

b ) The initial conditions

The initial conditions are : a_{0} = a_{1} = 0

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

7 0
3 years ago
What should be multiplied with -25/36 to get -5/9​
Ierofanga [76]

Answer:

\frac{4}{5}

Step-by-step explanation:

\frac{-\frac{5}{9}}{-\frac{25}{36}}=\\\\-\frac{5}{9}*(-\frac{36}{25})=\\\\\frac{4}{5}

6 0
2 years ago
What is the area of a sector with a central angle of 108 degrees and a diameter of 21.2 cm?
Rasek [7]

Answer:

105.84\ cm^2

Step-by-step explanation:

step 1

Find the area of the circle

The area of the circle is

A=\pi r^{2}

we have

r=21.2/2=10.6\ cm ----> the radius is half the diameter

substitute

A=\pi (10.6)^{2}

A=112.36\pi\ cm^2

step 2

Find the area of a sector

Remember that

The area of the circle subtends a central angle of 360 degrees

so

using proportion

Find out the area of a sector with a central angle of 108 degrees

\frac{112.36\pi}{360^o}=\frac{x}{108^o}\\\\x=112.36\pi(108)/360\\\\x=33.708\pi\ cm^2

assume

\pi=3.14

substitute

33.708(3.14)=105.84\ cm^2

7 0
3 years ago
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