4y + 10 = 5x 2.5x + 2y = -10

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

5 0

Answer:

are they seperate equations?

Step-by-step explanation:

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Length of a rectangle is 5 cm longer than the width. Four squares are constructed outside the rectangle such that each of the sq

stiv31 [10]

Answer:

Step-by-step explanation:

Remark

This is one of those questions that can throw you. The problem is that do you include the original rectangle or not. The way it is written it sounds like you shouldn't

However if you don't the question gives you 2 complex answers. (answers with the sqrt( - 1) in them.

Solution

Let the width = x

Let the length = x + 5

Area of the rectangle: L * w = x * (x + 5)

Area of the smaller squares (there are 2)

Area = 2*s^2

x = s

Area = 2 * x^2

Area of the larger squares = 2 * (x+5)^2

Total Area

x*(x + 5) + 2x^2 + 2(x + 5)^2 = 120 Expand

x^2 + 5x + 2x^2 + 2(x^2 + 10x + 25) = 120 Remove the brackets

x^2 + 5x + 2x^2 + 2x^2 + 20x + 50 = 120 collect the like terms on the left

5x^2 + 25x + 50 = 120 Subtract 120 from both sides.

5x^2 + 25x - 70 = 0 Divide through by 5

x^2 + 5x - 14 = 0 Factor

(x + 7)(x - 2) = 0 x + 7 has no meaning

x - 2 = 0

x = 2

Perimeter

P = 2*w + 2*L

w = 2

L = 2 + 5

L = 7

P = 2*2 + 2 * 7

P = 4 + 14

P = 18

8 0

3 years ago

Sin A tan A/ 1 - cos A = 1+ sec A

aalyn [17]

$\text{L.H.S}\\\\=\dfrac{\sin A \tan A}{1- \cos A}\\\\\\=\dfrac{\sin A \cdot \tfrac{\sin A}{\cos A} }{ 1- \cos A}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~;\left[\tan A = \dfrac{\sin A }{\cos A}\right]\\\\\\=\dfrac{\tfrac{\sin^2 A}{ \cos A}}{1- \cos A}\\\\\\=\dfrac{\sin^2 A}{\cos A(1-\cos A)}\\\\\\=\dfrac{1-\cos^2 A}{\cos A( 1- \cos A)}~~~~~~~~~~~~~~~~~~~~~~~~;[\sin^2 A + \cos^2 A = 1]\\\\\\$

$=\dfrac{(1+ \cos A)(1 - \cos A)}{\cos A(1 - \cos A)}\\\\\\=\dfrac{1+ \cos A}{ \cos A}\\\\\\=\dfrac{1}{\cos A} + \dfrac{\cos A}{ \cos A}\\\\\\=\sec A + 1 \\\\\\=1+ \sec A\\\\\\=\text{R.H.S}\\\\\text{Proved.}$