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3 years ago

The equation of a parabola is 116(y+3)2=x+4 . What are the coordinates of the focus? (0, −3) (−8, −3) (−4, −7) (−4, 1)

Mathematics

2 answers:

Temka [501]3 years ago

6 0

Did you get the answer yet

Vlada [557]3 years ago

6 0

Answer:

The answer is (0, -3). Just took test.

Step-by-step explanation:

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The distribution of the amount of money spent by first-time gamblers at a major casino in Las Vegas is approximately normal in s

Leya [2.2K]

Answer:

$720

Step-by-step explanation:

Given :

Mean, m = 600

Standard deviation, s = 120

Z = (x - m) / s

P(Z > x) = 84%

P(Z > x) = 0.84

Zscore corresponding to P(Z > x) = 0.84 will be -0.994

Hence,

-0.994 = (x - 600) / 120

120 * -0.994 = x - 600

119.28 = x - 600

119.28 + 600 = x

719.28 = x

Hence,

X = $720

4 0

2 years ago

Encuentre el angulo positivo y que es coterminal con -270

GalinKa [24]

Ok so ummmm idk how to speck in that but well ummmmm idk

6 0

2 years ago

- 7 and 8<br> What’s the answer

kvv77 [185]

Answer:

Step-by-step explanation:

8-7= 1

Hope this helps!

3 0

3 years ago

Solve the right triangle. Round decimal answers to the nearest tenth. find the sides and angles

babymother [125]

Answer:

b = 12.6

Step-by-step explanation:

a*2 + b*2 = c*2

6*2 + b*2 = 14*2

36 + b*2 = 196

Subtract 36 from both sides

36 + b*2 = 196

-36 -36

b*2 = 160

Find the square root of both

b = 12.64911064 or 12.6

4 0

3 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