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kipiarov [429]
3 years ago
11

PLZ HELPPPP

Mathematics
1 answer:
Zolol [24]3 years ago
6 0

Answer:

Step-by-step explanation:

3x-4<8

3x<12

x<4

2x+2>4

2x>2

>1

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What is the surface area of a sphere with a radius of 11 units
Otrada [13]

Answer: A≈1519.76³ Units

Step-by-step explanation:

A=4πr²

r = 11

11²=121

121 * 3.14 = 379.94

379.94(4) = 1519.76

A≈1519.76³ Units OR

A=4πr2=4·π·112≈1520.53084

4 0
3 years ago
There were 178 old cars at the junkyard. Then a truck brought in some more cars. Now there are a total of 236 cars in the junkya
crimeas [40]

Answer:

<u>The correct answer is 58 cars.</u>

Step-by-step explanation:

Cars in the junkyard before the truck arrived with some = 178

Cars after the truck brought into the junkyard some more= 236

Cars brought in by the truck = 236 - 178

Cars brought into the junkyard by the truck = 58

<u>The number of cars brought by the truck is 58 cars.</u>

5 0
3 years ago
a ship leave port p and sails on a bearing N50 degree E to Port Q 15km away.then it sails on a bearing of S45 degree E to Port R
Shtirlitz [24]

Answer:

Port r is 100° from Port p and 26km from Port p

Step-by-step explanation:

Lets note the dimension.

From p to q = 15 km = a

From q to r = 20 km= b

Angle at q = 50° + 45°

Angle at q = 95°

Ley the unknown distance be x

Distance from p to r is the unknown.

The formula to be applied is

X²= a²+ b² - 2abcosx

X²= 15² + 20² - 2(15)(20)cos95

X²= 225+400-(-52.29)

X²= 677.29

X= 26.02

X is approximately 26 km

To know it's direction from p

20/sin p = 26/sin 95

Sin p= 20/26 * sin 95

Sin p = 0.7663

P= 50°

So port r is (50+50)° from Port p

And 26 km far from p

3 0
3 years ago
Three listening stations located at (3300, 0), (3300, 1100), and (-3300, 0) monitor an explosion. The last two stations detect t
erastovalidia [21]

Answer:

The coordinates of the explosion is (3300, -2750)

Step-by-step explanation:

I have plot the points and attached to thus answer for easy understanding.

Now, from the question, since station A is the first to hear the explosion, we'll make it the foci of the parabola in the graph I attached and it will be horizontal since the distance between station C and A is much more than that between station B and A. Thus, the reason why station C will have to be the other foci with the hyperbola centred at the origin.

Now, sound travels at a speed of 1100 ft/s and station B is located 1100 ft from station A. Thus, the explosion would likely have occurred at a point on the line x = 3300ft . Since station A is 3300ft from centre C = 3300,hence C² = 3300² = 10,890,000. Since it takes 4 seconds longer for the sound to reach station C than A, the sound has traveled 4(1100)= 4400 ft.

Thus, 4400 = d1 = d2 = 2a

So,2a = 4400 and so, a =2200

a² = 2200² = 4,840,000 where d1 is the distance from station C to the explosion and d2 is the distance from station A to the explosion. To find b², let's use the equation ;

c² = a² + b² and so; b² = c² - a² = 10,890,000 - 4,840,000 = 6,050,000

Equation of hyperbola is given as;

(x²/a²) - (y²/b²) = 1

Plugging in the values of a² and b², we obtain ;

(x²/4,840,000) - (y²/6,050,000) = 1

Since we have deduced that the explosion must occur on the line x= 3300, we'll put in 3300 for x to obtain ;

(3300²/4,840,000) - (y²/6,050,000) = 1

2.25 - 1 = (y²/6,050,000)

y² = (1.25 x 6,050,000)

y² = 7562500

y = √7562500

y = ± 2750

Due to the fact that the explosion will occur at a point further from station B than from station A, the explosion will take place in quadrant 4. Thus, we will take the negative value of y which is - 2750.

So explosion will occur at the coordinate (3300, -2750)

7 0
3 years ago
What is the volume of a sphere with a radius of 25.6 cm, rounded to the nearest tenth of a cubic centimeter?
NARA [144]
You sure that’s the radius and not diameter? But if it is, the volume would be 70,276.2 cm
8 0
3 years ago
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