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3 years ago

Please assist me with these problems

Mathematics

1 answer:

Alex73 [517]3 years ago

8 0

Answer:

1) $350

2) 55°

3) $300

Step-by-step explanation:

1) The total cost is the cost of the appointment plus the cost of the repair work. The appointment costs $50. The repair work costs $120 per 2 hours. So the total cost is:

C = 50 + (120/2) t

C = 50 + 60t

For 5 hours of work, the cost is:

C = 50 + 60(5)

C = 350

It costs $350.

2) The temperature starts at 70°. It drops 10° in 4 hours. So the temperature after t hours is:

T = 70 + (-10/4) t

T = 70 − 2.5t

After 6 hours:

T = 70 − 2.5(6)

T = 55

The temperature is 55°.

3) Jennie's total pay is her weekly pay plus her commissions. If her commission is x% of her sales, then her pay is:

P = 250 + (x/100) S

When her sales is $1000, her pay is $275.

275 = 250 + (x/100) 1000

275 = 250 + 10x

25 = 10x

x = 2.5

So her pay is:

P = 250 + (2.5/100) S

When S = $2000:

P = 250 + (2.5/100) 2000

P = 250 + 50

P = 300

Her total pay is $300.

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vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

The Parry Glitter Company recently loaned $300,000 to FIX 92, a local radio station. The radio station signed a noninterest-bear

melomori [17]

Answer:

The Parry Glitter Company

The Parry Glitter Company should record the Notes Receivable as $300,000.

It should also record the interest receivable per year as $24,000 and the advertising cost as $24,000 per year. These bring into the accounting records the interest revenue and also the advertising expense, which eventually cancel each other.

Step-by-step explanation:

a) Data and Calculations:

Notes Receivable = $300,000

If the notes receivable are repaid at the end of 3 years and it is assumed that the interest on the notes receivable = 8%

Therefore, the cost of the free advertising will be equal to $24,000 ($300,000 * 8%), which is the cost of the interest to the radio station.

8 0

3 years ago

Mr. Chen filled a wading pool with 20 gallons of water. How many pints of water did he put in a pool

eduard

Answer:

160 pints

Step-by-step explanation: