Which sequence of transformations proves that shape I is similar to shape II?

Mathematics

2 answers:

4vir4ik [10]3 years ago

4 0

Note that if you simply reflected it and dilated it, it would be too far to the left

note that the distances bewteen the axises in image 1 were 1 and 3in image 3,the distances are 2 and 6
seems to be 3 times
so it's the 90 counterclockwise rotation and dilation by scale factor of 3
answer is last one

Travka [436]3 years ago

4 0

Answer:

The correct answer is C not D, because I got it right with C on PLATO.

Step-by-step explanation:

You might be interested in

a medium apple has 72 calories. This is 3/4 of the calories in a medium banana. How many calories does a medium banana have?

qaws [65]

The banana had 54 calories

8 0

3 years ago

Factor and solve X^2+7x-8=0

adelina 88 [10]

(X^2+8x)-(x-8)=0
x(x+8)-1(x+8)=0
(X-1)(x+8)=0
X= 1
X= -8

4 0

3 years ago

In the figure below, ray was constructed starting from rays and . By using a compass D and G were marked equidistant from E on r

DaniilM [7]

The correct answer is that the angles created are equal.

The steps given are used to create an angle bisector of the angle. That means it divides the angle into 2 congruent angles.

6 0

3 years ago

Which steps transform the graph of y = x2 to y = –2(x – 2)2 + 2? (1 point)

Kitty [74]

Y = x² to: y = - 2 ( x - 2 )² + 2
Answer:
D ) reflect across the x-axis, translate 2 units to the right, translate up 2 units, stretch by the factor of 2

7 0

3 years ago

If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =

attashe74 [19]

First of all, let θθ be some angle in (0,π)(0,π). Then

θθ is acute ⟺⟺ θ<π2θ<π2 ⟺⟺ cosθ>0cos⁡θ>0.θθ is right ⟺⟺ θ=π2θ=π2 ⟺⟺ cosθ=0cos⁡θ=0.θθ is obtuse ⟺⟺ θ>π2θ>π2 ⟺⟺ cosθ<0cos⁡θ<0.

Now, to see if (say) angle AA of the triangle ABCABC is acute/right/obtuse, we need to check whether cos∠BACcos⁡∠BAC is positive/zero/negative. But what is cos∠BACcos⁡∠BAC? It is the angle made by the vectors AB−→−AB→ and AC−→−AC→. (When you are computing the angle at a particular vertex vv, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex vv as the initial point.) We will first compute these two vectors:

AB−→−=(0,0,0)−(1,2,0)=(−1,−2,0)AB→=(0,0,0)−(1,2,0)=(−1,−2,0)AC−→−=(−2,1,0)−(1,2,0)=(−3,−1,0)AC→=(−2,1,0)−(1,2,0)=(−3,−1,0)Therefore, the angle between these vectors is given by:cos∠BAC=AB−→−⋅AC−→−|AB−→−||AC−→−|=…(1)(1)cos⁡∠BAC=AB→⋅AC→|AB→||AC→|=…Can you take it from here? From the sign of this value, you should be able to decide if angle AA is acute/right/obtuse.

Now, do the same procedure for the remaining two angles BB and CC as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

6 0

3 years ago