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Diano4ka-milaya [45]
3 years ago
5

SOMEONE HELP PLEASE

Mathematics
2 answers:
gulaghasi [49]3 years ago
8 0

Answer:

So the correct answer is option B- ( b - a )

Step-by-step explanation:

Given that M and N are the mid-points of RT and ST.

We are to find the length of MN.

As shown in the figure,

the co-ordinates of the point T are (2c, 2d),

the co-ordinates of the point S are (2b, 0),

and

the co-ordinates of the point R are (2c, 2d).

Since M is the mid-point of TR, so the co-ordinates of M are

\left(\dfrac{2c+2a}{2},\dfrac{2d+0}{2}\right)=(c+a,d).

Also, N is the mid-point of TS, the co-ordinates of N are

\left(\dfrac{2c+2b}{2},\dfrac{2d+0}{2}\right)=c+b,d)(.

Therefore, the length of the line segment MN calculated using distance formula will be

MN=\sqrt{(c+b-c-a)^2+(d-d)^2}=\sqrt{(b-a)^2}=b-a.

Thus, the required length of MN is (b - a) units.

Option (B) is correct.

yan [13]3 years ago
7 0
I have this same question, can anyone answer it?
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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.

8 0
3 years ago
Suppose Lin labels the books at a constant speed of s books per minute. There can be several equations written that represent th
Alla [95]

Answer:

S = \frac{b}{m}

Step-by-step explanation:

Given

Speed = s

Required

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The speed is calculated as:

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Substitute S for speed

S= \frac{Number\ of\ books}{Time}

Let the number of books be b and the time be m.

The expression becomes

S = \frac{b}{m}

Hence, the expression for the scenario is:

S = \frac{b}{m}

4 0
2 years ago
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