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3 years ago

14

Brett's retirement party will cost $14 if he invites 7 guests. If there are 10 guests, how much will Brett's retirement party co

st? Assume the relationship is directly proportional.

1 answer:

Fittoniya [83]3 years ago

5 0

She would have to spend $20 for the party (Ration 1:2) I hope this helped ^^

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Susan and terry run a day care center Since they use susans house it was agreed that her share is twice as Terrys if they earn 2

Nadya [2.5K]

The easiest way to do this is to divide 225 by 3, so 225 / 3, and what you get as the sum, in this case it's 75, you would multiply the sum by 2, which equals 150, and that, would be Susan's share. So... 225 / 3 = 75. 75 * 2 = 150. $150 is the amount Susan gets. Hope this helps!

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3 years ago

Please help me (not a test question)

lara [203]

Answer:

Yes

Step-by-step explanation:

4 x 3 = 12

5 x 3 = 15

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3 years ago

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What is the measures of angle ECD? Please show work.

bazaltina [42]

The anser is 40

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3 years ago

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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.

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3 years ago

18. Determine the common difference, the fifth term, and the sum of the first 100 terms of the following sequence:

Ksenya-84 [330]

$a_1=1,\ a_2=2.5,\ a_3=4,\ a_4=5.5,\ ...\\\\a_2-a_1=2.5-1=1.5\\a_3-a_2=4-2.5=1.5\\a_4-a_3=5.5-4=1.5\\a_{n+1}-a_n=1.5=constans\\\\\text{It's an arithmetic sequence with}\\a_1=1,\ \boxed{d=1.5}\\\\a_n=a_1+(n-1)d\to a_n=1+(n-1)(1.5)=1+1.5n-1.5\\\\a_n=1.5n-0.5\\\\a_5=1.5(5)-0.5=7.5-0.5=7\\\boxed{a_5=7}\\\\\text{The formula of a Sum of the First n Terms of an Arithmetric Sequence:}\\\\S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n\\\\\text{We have:}\\a_1=1,\ d=1.5,\ n=100\\\\\text{Substitute}$

$S_{100}=\dfrac{(2)(1)+(100-1)(1.5)}{2}\cdot100=\dfrac{2+(99)(1.5)}{1}\cdot50\\\\=(2+148.5)\cdot50=150.5\cdot50=7,525\\\\\boxed{S_{100}=7,525}\\\\Answer:\\the\ common\ difference:\ d=1.5\\the\ fifth\ term:\ a_5=7\\the\ sum\ of\ first\ 100\ terms:\ S_{100}=7,525$

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3 years ago