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Fed [463]
3 years ago
6

You are buying telephone equipment from a company whose prices are given in English Pounds currency. You need to purchase teleph

one equipment for 3 offices along with one complete switchboard system. Equipment for each office costs 100 Pounds, and the switchboard costs 333 Pounds. A 10% tax must also be added to the order. If the current exchange rate is $1.50 equals 1 Pound, how much will the system cost?
Mathematics
1 answer:
Fynjy0 [20]3 years ago
8 0
We will first calculate the total cost in pounds then convert to dollars.

1- Equipment for three offices. The cost is 100 pounds per equipment.
Total cost of equipment = 3*100 = 300 pounds ..........> I

2- One switchboard. The cost is 333 pounds.
Total cost of switchboard = 333 pounds ...............> II

3- Total cost excluding taxes can be calculated by adding I and II as follows:
Cost excluding taxes = 300 + 333 = 633 pounds

4- Tax is 10% of the cost.
Tax amount = 0.1 * 633 = 63.3 pounds

5- Total cost including taxes = 633 + 63.3 = 696.3 pounds

6- Convert the cost to dollars.
We know that 1 pound is equivalent to 1.5$. To know the equivalence of 696.3 pounds, just do cross multiplication as follows:
Cost in dollars = (696.3*1.5) / 1 = 1044.45$


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Evaluate integral _C x ds, where C is
borishaifa [10]

Answer:

a.    \mathbf{36 \sqrt{5}}

b.   \mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}

Step-by-step explanation:

Evaluate integral _C x ds  where C is

a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)

i . e

\int  \limits _c \ x  \ ds

where;

x = t   , y = t/2

the derivative of x with respect to t is:

\dfrac{dx}{dt}= 1

the derivative of y with respect to t is:

\dfrac{dy}{dt}= \dfrac{1}{2}

and t varies from 0 to 12.

we all know that:

ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \  \ dt

∴

\int \limits _c  \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt

= \int \limits ^{12}_{0} \  \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2})  \ dt

= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0

= \dfrac{\sqrt{5}}{4}\times 144

= \mathbf{36 \sqrt{5}}

b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)

Given that:

x = t  ; y = 3t²

the derivative of  x with respect to t is:

\dfrac{dx}{dt}= 1

the derivative of y with respect to t is:

\dfrac{dy}{dt} = 6t

ds = \sqrt{1+36 \ t^2} \ dt

Hence; the  integral _C x ds is:

\int \limits _c \ x \  ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \  dt

Let consider u to be equal to  1 + 36t²

1 + 36t² = u

Then, the differential of t with respect to u is :

76 tdt = du

tdt = \dfrac{du}{76}

The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145

Thus;

\int \limits _c \ x \  ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \  dt

\mathtt{= \int \limits ^{145}_{0}  \sqrt{u} \  \dfrac{1}{72} \ du}

= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}

\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}

\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}

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4 years ago
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pogonyaev

Answer:

- 3x + 10 =  - 20 \\  - 3x =  - 20 - 10 \\  - 3x =  - 30 \\  \frac{ - 3x}{ - 3}  =  \frac{ - 30}{ - 3}  \\  x = 10

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