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Aloiza [94]
3 years ago
8

The degree measure of one of two complementary angles is 30 less than twice that of the other. What are the degree measures of t

he angles? (Remember that complimentary angles add to be 90 degrees) 45 degrees and 45 degrees 30 degrees and 60 degrees 27 degrees and 63 degrees 40 degrees and 50 degrees
Mathematics
1 answer:
KatRina [158]3 years ago
3 0
Let n = one of the angles
Let 2n -30 = the other

n+ 2n -30 =90⁰  (because they are complementary)
Combine like terms --     3n -30 = 90
Add 30 to both sides of the equation
3n=120°
 Divide both sides by 3 and n=40°    2n -30 = 80 -30 = 50°
So one angle is 40° and the other is 50° and those angles are complementary. 
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Answer:

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Step-by-step explanation:

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Points scored in a basketball game vary directly with shots taken. If a player scores 50 points and takes 40
KATRIN_1 [288]

Answer:

She would have scored 75 points.

Step-by-step explanation:

Points scored in a basketball game vary directly with shots taken.

This means that this question can be solved by proportions, using a rule of three.

How many would she have scored if she takes 60 shots?

Using a rule of three:

50 points - 40 shots

x points - 60 shots

Applying cross multiplication:

40x = 50*60

Dividing both sides by 40

x = 50*1.5 = 75

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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)

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