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3 years ago

- Factorise 121y² - m²

Mathematics

1 answer:

Otrada [13]3 years ago

6 0

Answer:

$\boxed{ \bold{ { \boxed{ \sf{(11y + m)(11y - m)}}}}}$

Step-by-step explanation:

$\sf{121 {y}^{2} - {m}^{2} }$

$\longrightarrow{ \sf{ {(11y)}^{2} - {(m)}^{2} }}$

Use the formula of a² - b² = ( a + b ) ( a - b )

$\longrightarrow{ \sf{(11y + m)(11y - m)}}$

Hope I helped!

Best regards

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(d) A car travels at an average speed of 80 km/h. How far does it travel in: (1) 2 hours? (11) 12 minutes?

zepelin [54]

Answer is (1) 160 km in 2 hours and
(2) 16 miles in 12 minutes.

Step by step.
80 km / 1 hour = x km /2 hours
Cross multiply =
x = 160

The second one requires a little more math. Because we are looking at how far in minutes, we change the rate of 80km in 1 hour to 80 km in 60 minutes.
80 km / 60 = x / 12
Cross multiply
12x = 960
Solve for x by dividing both sides by 12
X = 16

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Mumz [18]

Answer:

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Step-by-step explanation:

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3 years ago

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Anton [14]

We have that
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<span>the actual area of a lake is 12 square miles

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the answer is
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4 years ago

Match each value of y with the corresponding value of the function?

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Answer:

The First one: f(-2) a. 9

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Help. <br>Please its urgent show workings.<br>

laiz [17]

Answer:

see explanation

Step-by-step explanation:

There are 2 possible approaches to differentiating these.

Expand the factors and differentiate term by term, or

Use the product rule for differentiation.

I feel they are looking for use of product rule.

Given

y = f(x). g(x) , then

$\frac{dy}{dx}$ = f(x).g'(x) + g(x).f'(x) ← product rule

(a)

y = (2x - 1)(x + 4)²

f(x) = 2x - 1 ⇒ f'(x) = 2

g(x) = (x + 4)²

g'(x) = 2(x + 4) × $\frac{d}{dx}$ (x + 4) ← chain rule

= 2(x + 4) × 1

= 2(x + 4)

Then

$\frac{dy}{dx}$ = (2x - 1). 2(x + 4) + (x + 4)². 2

= 2(2x - 1)(x + 4) + 2(x + 4)² ← factor out 2(x + 4) from each term

= 2(x + 4) (2x - 1 + x + 4)

= 2(x + 4)(3x + 3) ← factor out 3

= 6(x + 4)(x + 1)

--------------------------------------------------------------------------

(b)

y = x(x² - 1)³

f(x) = x ⇒ f'(x) = 1

g(x) = (x² - 1)³

g'(x) = 3(x² - 1)² × $\frac{d}{dx}$ (x² - 1) ← chain rule

= 3(x² - 1)² × 2x

= 6x(x² - 1)²

Then

$\frac{dy}{dx}$ = x. 6x(x² - 1)² + (x² - 1)³. 1

= 6x²(x² - 1)² + (x² - 1)³ ← factor out (x² - 1)²

= (x² - 1)² (6x² + x² - 1)

= (x² - 1)²(7x² - 1)

----------------------------------------------------------------------

(c)

y = (x² - 1)(x³ + 1)

f(x) = x² - 1 ⇒ f'(x) = 2x

g(x) = (x³ + 1) ⇒ g'(x) = 3x²

Then

$\frac{dy}{dx}$ = (x² - 1). 3x² + (x³ + 1), 2x

= 3x²(x² - 1) + 2x(x³ + 1) ← factor out x

= x[3x(x² - 1) + 2(x³ + 1) ]

= x(3x³ - 3x + 2x³ + 2)

= x(5x³ - 3x + 2) ← distribute

= 5 $x^{4}$ - 3x² + 2x

--------------------------------------------------------------------

(d)

y = 3x³(x² + 4)²

f(x) = 3x³ ⇒ f'(x) = 9x²

g(x) = (x² + 4)²

g'(x) = 2(x² + 4) × $\frac{d}{dx}$ (x² + 4) ← chain rule

= 2(x² + 4) × 2x

= 4x(x² + 4)

Then

$\frac{dy}{dx}$ = 3x³. 4x(x² + 4) + (x² + 4)². 9x²

= 12 $x^{4}$ (x² + 4) + 9x²(x² + 4)² ← factor out 3x²(x² + 4)

= 3x²(x² + 4) [ 4x² + 3(x² + 4) ]

= 3x²(x² + 4)(4x² + 3x² + 12)

= 3x²(x² + 4)(7x² + 12)

5 0

3 years ago

- Factorise 121y² - m²​

- Factorise 121y² - m²