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3 years ago

14

Pool is 25ft long and 15 ft wide and the width of the walkway is to be the same all around the pool. How wide do I make the walk

way?

1 answer:

EastWind [94]3 years ago

8 0

You would need a 4 foot wide sidewalk.

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Police Chase: A speeder traveling 40 miles per hour (in a 25 mph zone) passes a stopped police car which immediately takes off a

zubka84 [21]

Answer:

a. 18.34 s b. 327.92 m

Step-by-step explanation:

a. How long before the police car catches the speeder who continued traveling at 40 miles/hour

The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.

So, a = (v - u)/t = (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².

The distance moved by the police car in 10 s is gotten from

s = ut + 1/2at² where u = initial velocity of police car = 0 m/s, a = acceleration = 2.458 m/s² and t = time = 10 s.

s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²

s = 0 m + 1/2 × 2.458 m/s² × 100 s²

s = 122.9 m

The distance moved when the police car is driving at 55 mi/h is s' = 24.58 t where t = driving time after attaining 55 mi/h

The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t

The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' = 17.88t' m where t' = time taken for police to catch up with speeder.

Since both distances are the same,

S' = S

17.88t' = 122.9 + 24.58t

Also, the time taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t

t' = 10 + t

So, substituting t' into the equation, we have

17.88t' = 122.9 + 24.58t

17.88(10 + t) = 122.9 + 24.58t

178.8 + 17.88t = 122.9 + 24.58t

17.88t - 24.58t = 122.9 - 178.8

-6.7t = -55.9

t = -55.9/-6.7

t = 8.34 s

So, t' = 10 + t

t' = 10 + 8.34

t' = 18.34 s

So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour

b. how far before the police car catches the speeder who continued traveling at 40 miles/hour

Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m

4 0

2 years ago

Please help! Geometry

Vaselesa [24]

Hi :)
the answer is 16

7 0

2 years ago

45 points for first most detailed / accurate answer:

Bas_tet [7]

A.) Since there are no restrictions as to the dimensions of the candle except that their volumes must equal 1 cubic foot and that each must be a cylinder, we have the freedom to decide the candles' dimensions.

I decided to have the candles equal in volume. So, 1 cubic foot divided by 8 gives us 0.125 cubic foot, 216 in cubic inches.

With each candle having a volume of 216 cubic inches, I assign a radius to each: 0.5 in, 1.0 in, 1.5 in, 2.0 in, 2.5 in, 3.0 in, 3.5 in, and 4.0 in. Then, using the formula of the volume of a cylinder, which is:

V=pi(r^2)(h)

we then solve the corresponding height per candle. Let us let the value of pi be 3.14.

Hence, we will have the following heights (expressed to the nearest hundredths) for each of the radius: for

r=2.5 in: h=11.01 in
r=3.0 in: h= 7.64 in
r=3.5 in: h= 5.62 in
r=4.0 in: h= 4.30 in
r=4.5 in: h= 3.40 in
r=5.0 in: h= 2.75 in
r=5.5 in: h= 2.27 in
r=6.0 in: h= 1.91 in

b. each candle should sell for $15.00 each

($20+$100)/8=$15.00

c. yes, because the candles are priced according to the volume of wax used to make them, which in this case, is just the same for all sizes

7 0

2 years ago

An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then

Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

$\angle Q=110^\circ$

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

$q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km$

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

$=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)$

(b)Flight Direction of Y

Using Law of Sines

$\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)$

The direction of flight Y to the nearest degree is 39 degrees.

7 0

3 years ago

There are 15 red, 14 green, and 9 yellow balls in a dark room. What is the minimum number of balls that Alicia needs to take so

Tems11 [23]

Answer:

32

Step-by-step explanation:

1. If you take 12 balls, 9 could be yellow and 3 could be red. We want at least 12 of the same color, so 30 balls isn't a good minimum

2. If you take 30 balls, 9 could be yellow, 10 could be red and 11 could be green. We want at least 12 of the same color, so 30 balls isn't a good minimum.

3. If you take 31 balls, 9 could be yellow, 11 could be red and 11 could be green. We want at least 12 of the same color, so 31 balls isn't a good minimum.

4. If you take 32 balls, 9 could be yellow, 11 could be red and 12 could be green or 9 could be yellow, 12 could be red and 11 could be green or 7 could be yellow, 13 could be red and 12 could be green or 9 could be yellow, 11 could be red and 12 could be green.

We want at least 12 of the same color, and we have it from the different samples of the 4th trial.

So, 32 balls is the minimum number of balls

8 0

3 years ago