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Olenka [21]
3 years ago
11

4log, 11 – 20log 35

Mathematics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

  -26.71579

Step-by-step explanation:

Any scientific or graphing calculator can evaluate this expression for you.

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Step-by-step explanation:

11 degrees to radians: 1 degree =57.296 radians

L = (11 degrees)/(57.296) x (11 in) = 2.111848 in

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What is the volume of the cylinder below use formula v=ñr2 h
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Answer:

The volume of the cylinder = π r² h

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This is formula is applied for the right cylinder as figure 1 and oblique cylinder as figure2.

<u>The volume of the cylinder of figure 1:</u>

r = 6  and  h = 7

volume = π r² h = π  * 6² * 7 = 252π   units³

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The range of communication is about 55 meters.
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3 years ago
Simplify 2(12-18\3+4)
Goshia [24]

{\bold{\red{\huge{\mathbb{QUESTION}}}}}

Simplify

2(12 -  \frac{18}{3}  + 4)

\huge\green{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}}}

ATQ->

2(12 - ( \frac{ \cancel{18}^{ \:  \:  \: 6} }{3} ) + 4 \\ 2(12 - 6 + 4) \\ 2(16 - 6) \\ 2(10) \\ 20 \:  \: is \: your \: ans

4 0
3 years ago
Read 2 more answers
Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
soldier1979 [14.2K]

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0
3 years ago
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