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3 years ago

Parallelogram PQRS has diagonals that are perpendicular. Based on that, which statement is most correct?

Mathematics

2 answers:

viva [34]3 years ago

4 0

Answer:

B- Parallelogram PQRS is also a rhombus.

Step-by-step explanation:

Given

Parallelogram PQRS with perpendicular diagonals

Required

Which of the option is true

(a) PQRS can be a rectangle

A rectangle do not have perpendicular diagonals.

Hence, (a) is false

If (a) is false, then (d) is also false

(b) PQRS can a rhombus

The diagonals of a rhombus are not perpendicular.

So (b) is true

No need to check for (c), since only option is true

Sveta_85 [38]3 years ago

3 0

Answer:

B- Parallelogram PQRS is also a rhombus.

Step-by-step explanation:

Hope this helps

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Step-by-step explanation:

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dybincka [34]

i) The given function is

$f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}$

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$(x-5)(x+4)^2\ne 0$

$(x-5)\ne0,(x+4)^2\ne 0$

$x\ne5,x\ne -4$

ii) For vertical asymptotes, we simplify the function to get;

$f(x)=\frac{(2x+1)}{(x+4)^2}$

The vertical asymptote occurs at

$(x+4)^2=0$

$x=-4$

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

$2x+1=0$

$2x=-1$

$x=-\frac{1}{2}$

iv) To find the y-intercept, we substitute $x=0$ into the reduced fraction.

$f(0)=\frac{(2(0)+1)}{(0+4)^2}$

$f(0)=\frac{(1)}{(4)^2}$

$f(0)=\frac{1}{16}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0$

The horizontal asymptote is $y=0$ .

vi) The function has a hole at $x-5=0$ .

Thus at $x=5$ .

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

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