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3 years ago

In the figure, the shaded region is formed by removing a circle of centre P and radius 5 cm

Mathematics

1 answer:

S_A_V [24]3 years ago

7 0

$\large\boxed{\fcolorbox{red}{aqua}{Radius~of~semi~circle~=~10cm}}$

$\large\underline\mathrm\purple{Diameter~is~20cm}$

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3 years ago

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Dwayne drove 18 miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an av

Nesterboy [21]

Answer:

30 mph

Step-by-step explanation:

Let d = distance (in miles)

Let t = time (in hours)

Let v = average speed driving <u>to</u> the airport (in mph)

⇒ v + 15 = average speed driving <u>from</u> the airport (in mph)

Using: distance = speed x time

$\implies t=\dfrac{d}{v}$

Create two equations for the journey to and from the airport, given that the distance one way is 18 miles:

$\implies t=\dfrac{18}{v} \ \ \textsf{and} \ \ t=\dfrac{18}{v+15}$

We are told that the total driving time is 1 hour, so the sum of these expressions equals 1 hour:

$\implies \dfrac{18}{v} +\dfrac{18}{v+15}=1$

Now all we have to do is solve the equation for v:

$\implies \dfrac{18(v+15)}{v(v+15)} +\dfrac{18v}{v(v+15)}=1$

$\implies \dfrac{18(v+15)+18v}{v(v+15)}=1$

$\implies 18(v+15)+18v=v(v+15)$

$\implies 18v+270+18v=v^2+15v$

$\implies v^2-21v-270=0$

$\implies (v-30)(v+9)=0$

$\implies v=30, v=-9$

As v is positive, v = 30 only

So the average speed driving to the airport was 30 mph

(and the average speed driving from the airport was 45 mph)

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2 years ago

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