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Sati [7]
2 years ago
7

X

ign="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Sonbull [250]2 years ago
3 0

0.8265x = 45 \\ x = 45  \div 0.8265  \\ x = 45 \times  \frac{10000}{8265}  \\ x =  \frac{45000}{8265}  \\ x = 54 \frac{3690}{8265}  \\ x = 54 \frac{246}{551}

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713.49 written in a standard form
Vikki [24]

Answer: one hundred  seven thirteen and forty-nine hundreds

Step-by-step explanation: Standard form is a way of writing down very large or very small numbers easily. 103 = 1000, so 4 × 103 = 4000 . So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form.

7 0
1 year ago
A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si
Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

Area = 128

Let the dimension of the paper be x and y;

Such that:

Length = x

Width = y

So:

Area = x * y

Substitute 128 for Area

128 = x * y

Make x the subject

x = \frac{128}{y}

When 1 inch margin is at top and bottom

The length becomes:

Length = x + 1 + 1

Length = x + 2

When 2 inch margin is at both sides

The width becomes:

Width = y + 2 + 2

Width = y + 4

The New Area (A) is then calculated as:

A = (x + 2) * (y + 4)

Substitute \frac{128}{y} for x

A = (\frac{128}{y} + 2) * (y + 4)

Open Brackets

A = 128 + \frac{512}{y} + 2y + 8

Collect Like Terms

A = \frac{512}{y} + 2y + 8+128

A = \frac{512}{y} + 2y + 136

A= 512y^{-1} + 2y + 136

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

A' = -512y^{-2} + 2

Set

A' = 0

This gives:

0 = -512y^{-2} + 2

Collect Like Terms

512y^{-2} = 2

Multiply through by y^2

y^2 * 512y^{-2} = 2 * y^2

512 = 2y^2

Divide through by 2

256=y^2

Take square roots of both sides

\sqrt{256=y^2

16=y

y = 16

Recall that:

x = \frac{128}{y}

x = \frac{128}{16}

x = 8

Recall that the new dimensions are:

Length = x + 2

Width = y + 4

So:

Length = 8 + 2

Length = 10

Width = 16 + 4

Width = 20

To double-check;

Differentiate A'

A' = -512y^{-2} + 2

A" = -2 * -512y^{-3}

A" = 1024y^{-3}

A" = \frac{1024}{y^3}

The above value is:

A" = \frac{1024}{y^3} > 0

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

3 0
1 year ago
Jerry makes $40,000 a year working at a nearby factory. He gets two weeks paid vacation per year, plus five other paid holidays.
JulijaS [17]

Answer:

Vacation pays are not included in salaries. Therefore, Jerry's calculation is wrong.

Step-by-step explanation:

Given is :

Jerry makes $40,000 a year working at a nearby factory.

He gets two weeks paid vacation per year, plus five other paid holidays.

So total paid holidays become = 14+5=19 days

Subtracting 19 from 365 days and assuming that Jerry works for 365 days a year.

We get = 365-19=346 days

So, his per day salary will be = \frac{40000}{346}= 115.60

Vacation pays are not included in salaries. Therefore, Jerry's calculation is wrong.

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