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SVEN [57.7K]
4 years ago
13

Give the equation of the circle centered at the origin and passing through the point (4,0)

Mathematics
1 answer:
zimovet [89]4 years ago
4 0

Answer:

x² + y² = 4²

Step-by-step explanation:

The standard equation of a circle centered at the origin and with radius r is

x² + y² = r²

Note that the point (4, 0) is 4 units horizontally and to the right of the center (0, 0).  Thus, we know immediately that the radius is 4.

Thus the equation of this circle is x² + y² = 4²

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Solve for x. 8x2−3=2
gavmur [86]

8x^2 -3 = 2

add 3 to each side

8x^2 = 5

divide by 8

x^2 =5/8

take the square root of each side

x = +- sqrt (5/8)

7 0
3 years ago
Compute the permutations and combinations. 4 P 4<br> A) 1<br> B) 4<br> C) 24
Elenna [48]
I belive the answer would be B if not im sorry
4 0
4 years ago
A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th
Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

  • The bearing of the buoy from the is approximately <u>307.35°</u>

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}

Therefore;

\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}

Which gives;

\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}

Similarly, we can get;

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}

The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0
2 years ago
An architect is standing 250 feet from the base of a building and would like to know the height of the building. If he measures
GREYUIT [131]

see the figure below to better understand the problem

we have that

\begin{gathered} tan55^o=\frac{h}{250}---->\text{ by TOA} \\  \\ solve\text{ for h} \\ h=250*tan55^o \\ h=357.0\text{ ft} \end{gathered}<h2>The answer is 357.0 feet</h2>

8 0
1 year ago
The sum of angle 1 and angle 4 and the sum of angle 3 and angle 4 are each equal to 180 degrees by the definition of supplementa
S_A_V [24]

 Supplement Postulate? they are supplementary. 



4 0
3 years ago
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