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Phoenix [80]
3 years ago
8

How does a dilation with a scale factor of 1 affect the preimage

Mathematics
2 answers:
White raven [17]3 years ago
8 0
It keeps it the exact same. It is basically just multiplying everything by one, meaning it’s all the same just as if we did 4 times 1 which would equal 4.
zubka84 [21]3 years ago
6 0
I need to answer something it get points i hope you find the answers
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Please answer these for me !
levacccp [35]

Answer:

no i will not. hdjsjdjdidididididjjdjdjdiwkd

Step-by-step explanation:

DUMMY

5 0
3 years ago
A recent poll found that 35% of those surveyed are worried about aggressive drivers on the road. If three people are selected at
Ugo [173]

The probability that all three people surveyed is an independent probability and the calculated probability is 0.1225

The probability a surveyed driver is worried = 35%

  • P(worried) = 0.35

  • Number of people surveyed = 3

Since the probability of being worried is independent ;

  • Probability = P(A) × P(B) × P(C)

The probability that all 3 surveyed are worried :

  • P(worried) × P(worried) × P(worried)
  • (0.35 × 0.35 × 0.35) = 0.1225

Therefore, the probability that all three people surveyed are worried is : 0.1225

Learn more : brainly.com/question/18153040

5 0
2 years ago
What is an equation of the line that passes through the point (-4, -6) and is
const2013 [10]
Answer:y=2/3x-10/3
- 4/-6 is 4/6 so m is 4/6 then fill in the x and y variables in y=Mx+b solve for b then start plotting the equation of the line to check and you get your answer

I think this is the answer you might need to check again with more people
8 0
3 years ago
Read 2 more answers
All of the functions shown below are either exponential growth or decay functions.
AlekseyPX

Answer:

Step-by-step explanation:

If an exponential function is in the form of y = a(b)ˣ,

a = Initial quantity

b = Growth factor

x = Duration

Condition for exponential growth → b > 1

Condition for exponential decay → 0 < b < 1

Now we ca apply this condition in the given functions,

1). y=3.2(1+0.45)^{2x}

   Here, (1 + 0.45) = 1.45 > 1

   Therefore, It's an exponential growth.

2). y=(0.85)^{3x}

    Here, (0.85) is between 0 and 1,

    Therefore, it's an exponential decay.

3). y = (1 - 0.03)ˣ + 4

    Here, (1 - 0.03) = 0.97

    And 0 < 0.97 < 1

    Therefore, It's an exponential decay.

4). y = 0.5(1.2)ˣ + 2

    Here, 1.2 > 1

    Therefore, it's an exponential growth.

5 0
3 years ago
How do you solve this inequality; 1/3x&gt;-2?
Kisachek [45]
1/3x>-2=1>3x.-2
=1>-6x
=-1/6>x
4 0
3 years ago
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