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3 years ago

I dont know How to solve this at all

Mathematics

1 answer:

AlladinOne [14]3 years ago

7 0

So since this is a linear equation, which is y=mx+b, you use what it tells you in the word problem. so mx is the rate, since in this equation 200 per ton of sugar that's your mx, so its 200x. then for your b value, which in this case is the cost of truck rental, so its 5,400.

your equation is C=200s+5400

the variables are the same so you can use Y and X, but I used the variables used in the word problem.

Y=200x+5400

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3 years ago

In 1985, there were 285 cell phone subscribers in Arlington. The number of subscribers increased by 75% per year since 1985. Fin

ANTONII [103]

Answer:

The number of subscribers in 2008 is 110,845,988

Step-by-step explanation:

The number of cell phone subscribers in Arlington, in t years after 1985, can be modeled by the following equation.

$N(t) = N(0)(1+r)^{t}$

And which N(0) is the number of subscribers in 1985 and r is the rate it increases per year, as a decimal.

In 1985, there were 285 cell phone subscribers in Arlington.

This means that $N(0) = 285$

The number of subscribers increased by 75% per year since 1985.

This means that $r = 0.75$

$N(t) = 285(1.75)^{t}$

Find the number of subscribers in 2008.

2008 is 2008-1985 = 23 years after 1985. So we have to find N(23).

$N(23) = 285(1.75)^{23} = 110,845,988$

The number of subscribers in 2008 is 110,845,988

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3 years ago

Over summer, Jackie played video games 3 hours per day. When school begin in the fall, she was only allowed to play video games

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Answer:

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Step-by-step explanation:

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3 years ago

Stefan won 54 lollipops playing hoops at the county fair. At school he gave three to every student in his math class. He only ha

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3 years ago

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

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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

2 years ago