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4 years ago

Cora is using successive approximations to estimate a positive solution to

Mathematics

1 answer:

Advocard [28]4 years ago

7 0

Um...... what......?

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What additional information would prove that lmnp is a rectangle? the length of lm is startroot 45 endroot and the length of mn

cricket20 [7]

Rectangles are recognized to have the same qualities as parallelograms. The alternative with the additional information would demonstrate that LMNP is a rectangle is LP ⊥ PN.

<h3>What are rectangles?</h3>

A rectangle is a quadrilateral with four right angles in Euclidean plane geometry.
It can also be classified as an equiangular quadrilateral because all of its angles are equal, or a parallelogram with a right angle.
A square is a rectangle with four equal-length sides.

What to know about parallelogram and rectangle:

They both have two sorts of parallel sides as well as two pairs of opposite sides that are said to be congruent. All of the properties of a parallelogram are said to be shared by a rectangle.

This results in a rectangle and, invariably, a parallelogram.
However, a parallelogram is not generally referred to as a rectangle.

Option d, LP ⊥ PN as the supplementary information, would demonstrate that LMNP is a rectangle.

Therefore, rectangles are recognized to have the same qualities as parallelograms. The alternative with the additional information would demonstrate that LMNP is a rectangle is LP ⊥ PN.

Know more about rectangles here:

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The correct question is given below:
LMNP is a parallelogram.

What additional information would prove that LMNP is a rectangle?

A. The length of LM is √45 and the length of MN is √5.

B. The slope of LP and MN is –2.

C. LM ∥ PN

D. LP ⊥ PN

6 0

2 years ago

How to solve part ii and iii

iragen [17]

(i) Given that

$\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}$

when $x=1$ this reduces to

$\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}$

$\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}$

$2 \tan^{-1}(y) = \dfrac\pi3$

$\tan^{-1}(y) = \dfrac\pi6$

$\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)$

$\implies \boxed{y = \dfrac1{\sqrt3}}$

(ii) Differentiate $\tan^{-1}(xy)$ implicitly with respect to $x$ . By the chain and product rules,

$\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}$

(iii) Differentiating both sides of the given equation leads to

$\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0$

where we use the result from (ii) for the derivative of $\tan^{-1}(xy)$ .

Solve for $\frac{dy}{dx}$ :

$\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0$

$\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)$

$\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}$

$\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}$

$\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}$

From part (i), we have $x=1$ and $y=\frac1{\sqrt3}$ , and substituting these leads to

$\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}$