Compare the value of 30 and 3

Figure ABCD is reflected across the x-axis. What are the coordinates of A' , B' , C' , and D' ? Enter your answer in each box.

stealth61 [152]

Answer:

The coordinates of ABCD after the reflection across the x-axis would become:

A'(2, -3)

B'(5, -5)

C'(7, -3)

D'(5, -2)

Step-by-step explanation:

The rule of reflection implies that when we reflect a point, let say P(x, y), is reflected across the x-axis:

x-coordinate of the point does not change, but

y-coordinate of the point changes its sign

In other words:

The point P(x, y) after reflection across x-axis would be P'(x, -y)

P(x, y) → P'(x, -y)

Given the diagram, the points of the figure ABCD after the reflection across the x-axis would be as follows:

P(x, y) → P'(x, -y)

A(2, 3) → A'(2, -3)

B(5, 5) → B'(5, -5)

C(7, 3) → C'(7, -3)

D(5, 2) → D'(5, -2)

Therefore, the coordinates of ABCD after the reflection across the x-axis would become:

A'(2, -3)

B'(5, -5)

C'(7, -3)

D'(5, -2)

8 0

3 years ago

Factor completely x^8- 16.

Nataly_w [17]

Answer:

Step-by-step explanation:

So in this example we'll be using the difference of squares which essentially states that: $(a-b)(a+b)=a^2-b^2$ or another way to think of it would be: $a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$ . So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:

$x^8-16 = (x^4-4)(x^4+4)$

Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

$x^4-4 = (x^2-2)(x^2+2)$

So completely factored form is: $(x^2-2)(x^2+4)(x^4+4)$

I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because $x^2+4 = x^2-(-4)$ . and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: $x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i)$ just something that might be useful in some cases.