In the figure, and bisect each other. Complete the statements to prove that quadrilateral ABCD is a parallelogram.

Mathematics

1 answer:

Alisiya [41]3 years ago

5 0

Observe the figure below.

Statement Reason

1. AC and BD bisect each other Given

2. AE = EC and BE = ED Definition of bisection

3. $m \angle AEB = m \angle CED$ Vertical angle theorem

Vertical angle theorem states " When two lines intersect each other, the vertically opposite angles are always equal".

4. $\Delta ABE \cong \Delta CDE$ SAS criterion for congruence

5. $m \angle ACD = m \angle CAB$ Corresponding angles of congruent triangles are congruent

6. $AB \parallel CD$ Converse of alternate interior angle theorem

7. $m \angle BEC = m \angle AED$ Vertical angle theorem

8. $\Delta BEC \cong \Delta DEA$ SAS criterion for congruence

9. $BC \parallel AD$ Converse of alternate interior angle theorem

As, $\Delta BEC \cong \Delta DEA$ , so $m \angle CBD = m \angle BDA$ as corresponding parts of corresponding triangles are equal. As these angles are alternate interior angles, so the lines BC and AD are parallel by "Converse of alternate interior angle theorem".

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Which of the quadratic functions has the narrowest graph?

charle [14.2K]

Answer:

$y=\frac{1}{8}x^2$

Step-by-step explanation:

A quadratic function has the formula ax² + bx + c

To determine if a graph will be narrow or wide, the leading coefficient, a, will be the factor that determines this
The greater the coefficient, the narrower the parabola
The lesser the coefficient, the wider the parabola

Here all of the functions are in the form ax²

In $y=\frac{1}{8}x^{2}$ , our "a" term is $\frac{1}{8}$
In y = -2x², our "a" term is -2
In y = -3x², our "a" term is -3
In $y=\frac{1}{3}x^{2}$ , our "a" term is $\frac{1}{3}$

We can eliminate the two functions with the negative coefficients because they are much smaller than the two functions with the fractions as coefficients, and will therefore open much wider.

We can now compare the two remaining functions, $y=\frac{1}{8}x^2$ and $y=\frac{1}{3}x^2$

Giving the two fractions common denominators would turn them into $y = \frac{3}{24}x^2$ and $y=\frac{8}{24} x^2$
The equation with the larger fraction will be the parabola that is the narrowest. In this case, it is the $y=\frac{8}{24} x^2$ .
Therefore, $y=\frac{1}{8}x^2$ will have the narrowest graph