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Alona [7]
3 years ago
5

Is anyone good at this? Please help me!!

Mathematics
1 answer:
enyata [817]3 years ago
4 0
Speed: 30mi/sec
Seconds in one minute: 60
Seconds in 6 minute: 60 x 6 = 360 seconds
take the speed of 30miles / seconds and times it with 360, and you get a total distant traveled: 10800 miles
You might be interested in
Given the arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103,...
omeli [17]

We are given

arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103

so, first term is 124

u(1)= 124

now, we can find common difference

d=u(2)-u(1)

d=117-124

d=-7

now, we can find kth term

u(k)=u(1)+(k-1)d

now, we can plug values

and we get

u(k)=124+(k-1)*-7

u(k)=124-7k+7

u(k)=131-7k

u(k) must be negative

so,

u(k)=131-7k

131-7k

now, we can solve for k

7k>131

k>18.714

so, it's closest integer value is

k=19..............Answer


3 0
3 years ago
which of the circle diagrams below represnts the statement "if it is a square, then it is a quadrilateral
san4es73 [151]

Answer:

Step-by-step explanation:

yes

6 0
3 years ago
Which angle corresponds to 1/6 of a circle?
Basile [38]
360/6 = 60 degrees.

2pi/6 and pi/3 makes sense.

Out of 1/6 = 60 there's your answer.

Brainliest please!
5 0
3 years ago
Read 2 more answers
Prove: cot x (tan x + sin x) =1+ cos x
Sergio039 [100]
I hope this helps you

3 0
3 years ago
Plz hurry!!!! thank you!!!!
Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = \frac{115}{2} = 57.5°

Now, tan(57.5°) = \frac{OS}{SN}

⇒ 1.5697 = \frac{R}{SN}

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = \frac{NQ}{QM}

⇒ tan65° = \frac{NQ}{QM}

⇒ QM = \frac{2R}{2.1445}

QM = 0.9326 R .

⇒ MT = MQ + QT

          = 0.9326 R + 0.637 R  (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = (\frac{NP + MK}{2}) × (ST)

       = (\frac{1.274 R + 3.1392 R}{2}) × 2 R

       = 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0
3 years ago
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