If a shape is translated (has it's vertex' coordinates moved/changed), then it retains its original values but just has different vertex coordinates.

Thus since DE is equal to UV, then,

DE = UV; substitution:
5=5

5 0

4 years ago

Use the completing the square method to find the roots of 2x^2+3x-15=13 include the steps, it would be greatly appreciated thank

ValentinkaMS [17]

Answer:

The roots are

$x1= \frac{-3+\sqrt{233}}{4}$

$x2= \frac{-3-\sqrt{233}}{4}$

Step-by-step explanation:

we have

$2x^{2}+3x-15=13$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$2x^{2}+3x=13+15$

$2x^{2}+3x=28$

Factor the leading coefficient

$2(x^{2}+(3/2)x)=28$

Complete the square. Remember to balance the equation by adding the same constants to each side

$2(x^{2}+(3/2)x+(9/16))=28+(9/8)$

$2(x^{2}+(3/2)x+(9/16))=233/8$

$(x^{2}+(3/2)x+(9/16))=233/16$

Rewrite as perfect squares

$(x+(3/4))^{2}=233/16$

square root both sides

$(x+\frac{3}{4})=(+/-)\sqrt{\frac{233}{16}}\\ \\(x+\frac{3}{4})=(+/-)\frac{\sqrt{233}}{4}\\ \\x= -\frac{3}{4}(+/-)\frac{\sqrt{233}}{4}$

$x1= -\frac{3}{4}(+)\frac{\sqrt{233}}{4}=\frac{-3+\sqrt{233}}{4}$

$x2= -\frac{3}{4}(-)\frac{\sqrt{233}}{4}=\frac{-3-\sqrt{233}}{4}$

6 0

3 years ago