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ser-zykov [4K]
3 years ago
15

Suppose you are offered a job that lasts one month, and you are to be very well paid. Which of the following methods of payment

is more profitable for you?
Mathematics
1 answer:
mihalych1998 [28]3 years ago
7 0

Question is Incomplete, Complete question is given below.

Suppose you are offered a job that lasts one month. Which of the following methods of payment is more profitable for you?

I. One million dollars at the end of the month.

II. One cent on the first day of the month, two cents on the second day, four cents on the  third day, and, in general,  cents on the nth day.

Answer:

Option II is more profitable.

Step-by-step explanation:

Solution:

By Option I:

Amount we will get every month = $1,000,000

Now By Option II.

Given:

Amount we will get every month = 2^{n-1}

Now we will consider the month which has least number of days in a month which is February which has 28 days.

So we can say that.

Amount we will get every month = 2^{27-1}=2^{26}= 1, 34,217,728\ cents = \$1,342,177.28

$1,342,177.28 > $1,000,000

Hence Option II is more profitable.

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(Full question above)
zlopas [31]

Answer:

B. 300

Step-by-step explanation:

Final figure volume = big cuboid volume - small cuboid volume

Volume of a cuboid = height x width x length

Final figure = (6 x 6 x 15) - (4 x 4 x 15)

Final figure = 540 - 240

Final figure = 300mm^2

5 0
3 years ago
What is the measure of angle RSL if it is (9x+27)° and angle BSA is (6x+66)°
Shalnov [3]

Answer:

Assume:

(angle) RSB and angle (LSA) = D

and (line) ASR and (line) LSB is striaght then

(9x +27) + D = 180

(6x + 66) + D = 180

9x + D = 153

6x + D = 114

substitute

6x + D = 114

D = 114 - 6 x

9x + (114 - 6x) =153

3x = 39

x = 13

D = 114 - 6(13)

D=36


Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
Write the equation of the line that passes through the given points:
Airida [17]

Answer:

y+1=5/2(x+2)

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-6-(-1))/(-4-(-2))

m=(-6+1)/(-4+2)

m=-5/-2

m=5/2

y-y1=m(x-x1)

y-(-1)=5/2(x-(-2))

y+1=5/2(x+2)

4 0
3 years ago
Draw a picture to show the product of 4:7 x 3. Then write the product
Hoochie [10]

Answer:

¹²/₇

Step-by-step explanation:

Model ⁴/₇ × 3

Assume that you have three pies, each dived into seven slices.

There are only four slices remaining in each pie, that is, there is ⁴/₇ pie in each pie plate

The picture represents ⁴/₇ × 3.  

Model the product

Now, transfer slices to get as many filled pie plates as possible.

Count the total slices.

You have 12 slices, and each slice represents ⅐ of a pie.

You have ¹²/₇  pie.

∴ ⁴/₇ × 3 = ¹²/₇

6 0
3 years ago
What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?
scoray [572]

Answer:

<u />\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:
\displaystyle \lim_{x \to c} x = c

Special Limit Rule [L’Hopital’s Rule]:
\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:
\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]
Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given limit</em>.

\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}

<u>Step 2: Find Limit</u>

Let's start out by <em>directly</em> evaluating the limit:

  1. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    \displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}
  2. Evaluate:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

  1. [Limit] Apply Limit Rule [L' Hopital's Rule]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}
  2. [Limit] Differentiate [Derivative Rules and Properties]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}
  3. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}
  4. Evaluate:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}

∴ we have <em>evaluated</em> the given limit.

___

Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

___

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

3 0
1 year ago
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