Would it be A, B ,C or D
1 answer:
Answer:
D. statement is true.
Step-by-step explanation:
Given:
Δ DEF was dilated by a scale factor 3 to create Δ D'E'F'
Solution:
A dilation is a transformation that produces an image that is the same shape as the original, but is of different size.
So, the shape of the triangles DEF and D'E'F' are the same.
Since the shape is similar the angle of the dilated triangle and considered triangle are the same.
A. Segment DF ≅ Segment D'F'
This statement is false as the triangles which are dilated by scale factor 3 cannot be congruent to each other
B.
This statement is false since the shape is similar the angle of the dilated triangle and considered triangle are the same.
C. Segment EF ≅ 3. Segment E'F'
This statement is false as dilation is a transformation that produces an image that is the same shape as the original, but is of different size so it cannot be congruent.
D.
This statement is True as dilation is a transformation that produces an image that is the same shape as the original,Since the shape is similar the angle of the dilated triangle and considered triangle are the same.
Hence Option D. is True
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Answer:
x = 2
Step-by-step explanation:
These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.
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<h3>Squaring</h3>The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.
The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.
x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true
x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution
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<h3>Substitution</h3>Another way to solve this is using substitution for one of the radicals. We choose ...
Solutions to this equation are ...
u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution
The value of x is ...
x = u² -2 = 2² -2
x = 2 . . . . the solution to the equation
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<em>Additional comment</em>
Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.
