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babunello [35]
3 years ago
6

Figure ABCD is transformed to figure A prime B prime C prime D prime, as shown: A coordinate grid is shown from negative 5 to 0

to 5 on both x -and y-axes. A polygon ABCD has A at ordered pair 1, 3, B at ordered pair 3, 4, C at ordered pair 2, 1, D at ordered pair 1, 1. A polygon A prime B prime C prime D prime has A prime at ordered pair negative 3, 3, B prime at ordered pair negative 5, 4, C prime at ordered pair negative 4, 1, D prime at ordered pair negative 3, 1. Which of the following sequences of transformations is used to obtain figure A prime B prime C prime D prime from figure ABCD? (1 point) Reflection about the x-axis followed by a translation to the right by 2 units Reflection about the y-axis followed by a translation to the left by 2 units Counterclockwise rotation by 90 degrees about the origin followed by a translation to the right by 2 units Counterclockwise rotation by 90 degrees about the origin followed by a translation to the left by 2 units
Mathematics
1 answer:
Bess [88]3 years ago
7 0

Answer:

i need this answer too

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Answer:

f(x)=(x-1)^2+5 with domain x>1 and range y>5 has inverse g(x)=sqrt(x-5)+1 with domain x>5 and range y>1.

Step-by-step explanation:

The function is a parabola when graphed. It is in vertex form f(x)=a(x-h)^2+k where (h,k) is vertex and a tells us if it's reflected or not or if it's stretched. The thing we need to notice is the vertex because if we cut the graph with a vertical line here the curve will be one to one. So the vertex is (1,5). Let's restrict the domain so x >1.

* if x>1, then x-1>0.

* Also since the parabola opens up, then y>5.

So let's solve y=(x-1)^2+5 for x.

Subtract 5 on both sides:

y-5=(x-1)^2

Take square root of both sides:

Plus/minus sqrt(y-5)=x-1

We want x-1>0:

Sqrt(y-5)=x-1

Add 1 on both sides:

Sqrt(y-5)+1=x

Swap x and y:

Sqrt(x-5)+1=y

x>5

y>1

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