Drag each tile to the table to multiply
(6x – y)(2x – y + 2).

Step-by-step explanation:

Given :-

The length of the garden 8m greater than 2 times the width.

Area of the garden is 280 m²

Let us consider the length as x and width as y.

Sp, we can day length as :-

x = 8 + 2y ---(1)

Now, we know that:-

Area of Rectangle = Length × Breadth

280 = x * y

We can replace the value of x now,

280 = y × ( 8 + 2y)

280 = 8y + 2y²

2y² + 8y - 280 = 0

y² + 4y - 140 = 0

Factorise it.

(y -10)(y + 14)

Cancelling -ve value, we get the width as 10 metres.

Hope it helps :)

Answer:

Step-by-step explanation:

Width = w

Length = 2w + 8

Area of rectangular garden = 280 square meter

length * width = 280

(2w + 8 ) *w = 280

2w * w + 8*w = 280

2w² + 8w = 280

2w² + 8w - 280 = 0

Divide the whole equation by 2

w² + 4w - 140 = 0      

w² + 14w - 10w - 14 *10 = 0

w(w + 14) - 10(w + 14) = 0

(w + 14)(w - 10)= 0

w - 10 = 0       {Ignore w + 14, as measurements will not be -ve}

w = 10 m

l = 2*10 +8

= 20 +8

l  = 28 m

Answer:

4

Step-by-step explanation:

[tex] Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x} [/tex]

[tex] =Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x}\times \frac{1+\cos x}{1+\cos x} [/tex]

[tex] =Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{1^2 -\cos^2 x} [/tex]

[tex] =Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{1 -\cos^2 x} [/tex]

[tex] =Lim_{x \to 0}\frac{2 x\sin x(1+\cos x) }{sin^2 x} [/tex]

[tex] =Lim_{x \to 0}\frac{2x(1+\cos x) }{sin x} [/tex]

[tex] =Lim_{x \to 0} 2(1+\cos x) \times \frac{1}{Lim_{x \to 0}\frac{sin x}{x}} [/tex]

[tex] =2(1+\cos 0) \times 1 [/tex]

[tex] = 2(1+1) [/tex]

[tex] = 2(2) [/tex]

[tex] \therefore Lim_{x \to 0}\frac{2 x\sin x}{1-\cos x}= 4 [/tex]