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aalyn [17]
3 years ago
10

For what value of b will f(x)= x^2 + bx + 900 have -30 as its only zero? PLEASE HELP!!

Mathematics
1 answer:
kiruha [24]3 years ago
7 0

Answer:

60

Step-by-step explanation:

If -30 is its only zero, than the factors are (x+30)(x+30) = x^2 + 60x + 900

Than b = 60

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The diagonals of rhombus fghj intersect at point k. if side gh is equal to 3x - 10 and side jh is equal to 6x - 19, find x
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A rhombus is a square but turned so it looks like a diamond. This means all the sides are equal. With this information we can build an equation to solve for x.

3x - 10 = 6x - 19

1) Subtract 3x from both sides.

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2) Add 19 to both sides.

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3) Divide both sides by 3.

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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

    = 5x^{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

    = 12x^{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
2 years ago
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