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nadezda [96]
3 years ago
9

What is the image of (6,-2) under the translation (x,y) (x-5,y-8)

Mathematics
1 answer:
german3 years ago
6 0
The new point would then be at (1, -10)
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Six less than a number is twenty
nignag [31]

Answer:

the start numbew woould be 26

Step-by-step explanation:

6 less than 26 is 20

6 0
2 years ago
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Use the figure to find n please.
ira [324]

Answer:

n = 5

Step-by-step explanation:

Since this is a right triangle, we can use trig

tan theta = opp /adj

tan 30 = n / 5 sqrt(3)

5 sqrt(3) tan 30 = n

5 sqrt(3) * 1/ sqrt(3) = n

5 = n

4 0
2 years ago
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A police department commissioned a door-to-door survey of citizens on whether they commit minor law infractions, such as speedin
Neko [114]

Answer: Interviewer-induced bias

Step-by-step explanation:

The example given in the question illustrates the Interviewer-induced bias. This occurs when the situation created by the interviewer is such that the respondent will have to give the kind f answer that the interviewer wants to hear.

For example, in the question, we can deduce that the interviewers told the interviewees that they won't be investigated and the gave bias replies based on that.

4 0
2 years ago
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
Help please :( I can’t seem to get the answer
igomit [66]

Answer:

6 i don't know what the unit is, but its 6 for time

Step-by-step explanation:

It is a two step problem

54=9t

divide both sides by 9 to get the value of t

6=t

I hope that helps you :)

4 0
3 years ago
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