Respuesta :

Answer:

1) C. 4 - 3·i

2) D. The second graph shares the same vertex, is inverted, and opens wider than the first graph

3) C. y = (x - 2)² + 3 would shift right two units

4) B. Figure B' is congruent but not similar to figure B

5) A. m∠k' = m∠k

Step-by-step explanation:

1) Given that the real part of the complex number = 4

The imaginary of the complex number = -3

The general form of representing complex numbers is z = a + b·i, we have;

The binomial equivalent to the complex number is z = 4 - 3·i

2) The first graph equation is y = 2·x²

When x = 1, y = 2 and when x = 2, y = 8

The vertex = (h, k)

Where;

h = -b/(2.a) and b = 0, a = 2

∴ h = 0/(2 × 2) = 0

h = 0

k = f(h) = f(0) = 2 × 0² = 0

k = 0

The vertex, (h, k) = (0, 0)

The coefficient, 'a' is positive, therefore, the graph opens down

The second function, y = -(1/2)·x² also has a vertex (h, k) = (0, 0)

The coefficient, 'a' is positive, therefore, the graph opens up

When x = 1, y = -1/2 and when x = 2, y = -2

Therefore, the second function is wider

Therefore;

The second graph shares the same vertex, is inverted, and opens wider than the first graph

3) The given functions are;

First function; y = x² + 3 and second function; y = (x - 2)² + 3

First function;

When x = 1, y = x² + 3 = 1 + 3 = 4

∴ When x = 1, y = 4

Second function;

When y = 4, y = 4 = (x - 2)² + 3

√(1) = x - 2

x = 3

∴ When x = 3, y = 4

First function;

When x = 2, y = x² + 3 = 4 + 3 = 7

∴ When x = 2, y = 7

Second function;

When y = 7, y = 7 = (x - 2)² + 3

√4 = 2 = (x - 2)

x = 2 + 2 = 4

x = 4

∴ When x = 4, y = 7

Therefore, the second function, y = (x - 2)² + 3, has the x-value shifted 2 units to the right for a given value of 'y'

4) The lengths of the sides of figure B are 3 by 4, the lengths of the sides of figure B' 4.5 by 6

The ratio of the corresponding length and width of figures B and B' are;

3/4.5 = 4/6

Therefore, figure B' is similar but not congruent to figure B

5) A rotation and a reflection are rigid transformations and therefore, the dimensions and measure of the original figure and the image are the same;

∴ m∠k' = m∠k.