${9}^{ \frac{3}{2} } \times {27}^{ \frac{1}{2} } = {(3)}^{ \frac{3}{2} \times 2 } \times {(3)}^{ \frac{1}{2} \times 3} \\ = {3}^{3} \times {3}^{ \frac{3}{2} } \\ = {3}^{3 + \frac{3}{2} } \\ = {3}^{4\frac{1}{2} } \\ = {3}^{ \frac{9}{2} }$

8 0

2 years ago

Rewrite the equation by completing the square. 4 x^{2} -12 x +9 = 04x 2 −12x+9=0

DedPeter [7]

Answer:

(2x -3)^2 = 0

Step-by-step explanation:

In this particular instance, the square is already complete. The rewrite is to show that.

(2x -3)^2 = 0

In the general case, ...

(ax +b)^2 = a^2x^2 +2abx +b^2

Comparing this form to the given left-side expression, we see that ...

a = 2, b = -3

So, the trinomial is already a perfect square trinomial.

4 0

3 years ago

The coordinates of the vertices of a rectangle are (-4, 3), (-1, 5),(3, – 1), and (0, – 3).

SCORPION-xisa [38]

Answer:

Perimeter of rectangle is 23.34

Step-by-step explanation:

The coordinates of the vertices of a rectangle are (-4, 3), (-1, 5),(3, – 1), and (0, – 3).

What is the perimeter of the rectangle?

The formula to find perimeter of rectangle is: $Perimeter=2(l+w)$

We need to find length and width first using the coordinates (-4, 3), (-1, 5),(3, – 1), and (0, – 3).

Let A=(-4, 3) B=(-1, 5) C=(3, – 1) and D=(0, – 3) according to the figure attached.

So, If we find distance of point AB it can be considered length of rectangle and distance of point AC can be considered width of rectangle

The formula used to find distance is: $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Finding distance AB, we have $x_1=-4, y_1=3, x_2=-1, y_2=5$

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(-1-(-4))^2+(5-3)^2}\\Distance=\sqrt{(-1+4)^2+(5-3)^2}\\Distance=\sqrt{(3)^2+(2)^2}\\Distance=\sqrt{9+4}\\Distance=\sqrt{13}\\Distance=3.61$

So distance between point A and B is 3.61

Finding distance AC, we have $x_1=-4, y_1=3, x_2=3, y_2=-1$

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(3-(-4))^2+(-1-3)^2}\\Distance=\sqrt{(3+4)^2+(-1-3)^2} \\Distance=\sqrt{(7)^2+(-4)^2} \\Distance=\sqrt{49+16} \\Distance=\sqrt{65}\\Distance=8.06$

So, distance between point A and C is 8.06

We have length = 3.61 and width = 8.06

Using formula of perimeter

$Perimeter=2(l+w)\\Perimeter=2(3.61+8.06)\\Perimeter=2(11.67)\\Perimeter=23.34$

So, Perimeter of rectangle is 23.34

6 0

3 years ago