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2 years ago

HELLLLLLLLLLLLLPPPPPPPPPP

Mathematics

2 answers:

Liula [17]2 years ago

6 0

Answer:

Step-by-step explanation:

mark me brainliest pl

lawyer [7]2 years ago

3 0

The first option. 100% correct.

$a_n = 3\cdot a_{n-1}\\\\a_1 = 7\\\\a_2 = 3 a_{2-1} =3a_1 = 3(7) =21\\\\a_3 = 3 a_{3-1} =3a_2 = 3(21) = 63 \\\\a_4 = 3 a_{4-1} = 3 a_3 = 3 (63)=189\\\\7,21,63,189,......$

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One survey estimates that,on average, the retail value of a mid sized car decreases by8% annually.if the retail value of a car i

Scrat [10]

Percentage by which the average value of mid sized car decreases each year = 8%
Retail value of a car today = v dollars
Amount of decrease in the value of the car after 1 year = (8/100) * v
   = 2v/25 dollars
Then
The equation that represents the value of the car after 1 year = v - (2v/25) dollars
   = (25v - 2v)/25 dollars
   = 23v/25 dollars
So following this expression the value of the mid sized car can be easily determined after 1 year. I hope this is the answer you were looking for and the procedure is also clear to you.

5 0

3 years ago

Read 2 more answers

Help a girl out plzz:( im confused

nikdorinn [45]

Answer:

Step-by-step explanation:

the quotient means divide

7/y

the first number is the numerator and the second number is the denominator

7 0

2 years ago

Someone please help me!

bija089 [108]

Answer:

A = 58

B =:25

Step-by-step explanation:

C = A + B

83 = 9x + 4 + 4x + 1

83 = 13x + 5

13x = 78

x = 78/13

x = 6

A = 9x + 4 = 9 x 6 + 4 = 58

B = 4x + 1 = 4 x 6 + 1 = 25

7 0

2 years ago

Please help me with this question

boyakko [2]

Answer:

0.7999989281

Step-by-step explanation:

x1 = cos^-1 (3/5)

x1= 53.13°

sin(53.13) = 0.7999989281

5 0

3 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago