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3 years ago

The table below shows the outputs y for different inputs x:

Mathematics

2 answers:

vovangra [49]3 years ago

4 0

<h2>Answer with explanation:</h2>

Part A:

A relation is said to be a function where each input has only a single output.

We usually take the input variable as 'x' and the output variable as 'y'.

In the given table, we can see that 1 is the input value corresponding to 4 and 2 both.

i.e. one input value corresponds to two different output values that contradicts the definition of the function.

Thus the given table doesn't represent any function.

Part B: Given : The total cost f(x), in dollars, for renting a paddle-boat for x hours is shown below:

$f(x) = 20 + 10x$

At x= 120 , $f(120) = 20 + 10(120)=20+1200=1220$

It represents that the total cost for renting a paddle-boat for 120 hours is $1220.

igor_vitrenko [27]3 years ago

3 0

Part A)
The table doesn't represent y as a function of x.
Why? <span>One reason is that 1 or 3 is the first element in more than one ordered pair.
Part B)
f(x) = 10x + 20
f(120)= 10*120 + 20 = 1200 + 20 = 1220
</span><span>f(120) represents cost of boating for 120 hours.</span><span>
</span>

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Answer: $\$107.82$

Step-by-step explanation:

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3 years ago

How do you write y+1=4/5(x-3) in standard form?

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Answer:

The standard form of the given expression is $- (\frac{4}{5})x + y + (\frac{17}{5}) = 0$

Step-by-step explanation:

Here, the given expression is $y + 1 = \frac{4}{5} (x-3)$

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Solving this, we get:

$y + 1 = \frac{4}{5} (x-3) \implies y + 1 = (\frac{4}{5})x - \frac{12}{5}$

⇒ $y + 1 - (\frac{4}{5})x + (\frac{12}{5}) = 0$

or, $y - (\frac{4}{5})x + (\frac{12}{5} + 1 )= 0 \implies y - (\frac{4}{5})x + (\frac{12+ 5}{5}) = 0$

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4 years ago

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3 years ago

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A 20-ft by the 40-ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway i

Jlenok [28]

<h3>Answer: 2 feet</h3>

========================================================

Explanation:

x = width of the walkway in feet

This is some positive real number.

The dimension of 20 feet bumps up to 20+2x when adding on x from both directions. Similarly, the 40 ft dimension becomes 40+2x

Refer to the diagram below.

The 20 ft by 40 ft pool is surrounded by a larger rectangle that is 20+2x ft by 40+2x ft

The pool itself is 20*40 = 800 sq ft. Add on the walkway area to get 800+256 = 1056 sq ft.

-----------

Area = length*width

1056 = (20+2x)*(40+2x)

1056 = 20*40 + 20*2x + 2x*40 + 2x*2x ... FOIL rule

1056 = 800 + 40x + 80x + 4x^2

0 = 4x^2 + 40x + 80x + 800 - 1056

0 = 4x^2 + 120x - 256

4x^2 + 120x - 256 = 0

4(x^2 + 30x - 64) = 0

x^2 + 30x - 64 = 0

Let's use the quadratic formula to finish solving for x.

Plug in a = 1, b = 30, c = -64

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-30\pm\sqrt{(30)^2-4(1)(-64)}}{2(1)}\\\\x = \frac{-30\pm\sqrt{1156}}{2}\\\\x = \frac{-30\pm34}{2}\\\\x = \frac{-30+34}{2} \ \text{ or } \ x = \frac{-30-34}{2}\\\\x = \frac{4}{2} \ \text{ or } \ x = \frac{-64}{2}\\\\x = 2 \ \text{ or } \ x = -32\\\\$

Recall we made x be positive. This is because a negative walkway width does not make sense. This means we'll ignore x = -32.

The only practical solution is x = 2

Therefore, the walkway is <u>2 feet wide</u>

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