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2 years ago

What is the next term in the series -27,-18,-12,-8,...?

2 answers:

7 0

The next term in the series is 5 1/3.

The geometric ratio is 2/3. To find r you when to divide f(2) by f(1) and then you get 2/3. Then you use a formula called the explicit rule which is f(n)=f(1) • r^n-1.

Plug in the numbers. f(5)=-27•(2/3)^4

Then you get 5 1/3.

vlada-n [284]2 years ago

3 0

Answer:

We have - 18 / - 27 = 2 / 3;

- 12 / - 18 = 2 / 3;

-8 / - 12 = 2 / 3;

Then, x / - 8 = 2 / 3 ;

Finally, x = - 16 / 3;

The next terms in that series is - 16 / 3 .

Step-by-step explanation:

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Please help.

Charra [1.4K]

Answer:

Since we have a 30° - 60° - 90° triangle, we can calculate any side by knowing at least one out of three:

Since the length of the hypotenuse is twice as long as the shorter leg, we have:

The length of the short leg is: 25/2 = 12.5

Since the length of the longer leg is equal to the length of the shorter leg

multiply by square root of 3 we have:

The length of the longer leg is: 12.5 × √3 ≈21.65

=> The perimeter of the triangle is: 25 + 12.5 + 21.65 = 59.15

6 0

2 years ago

Read 2 more answers

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0

3 years ago

Bobby knows that the perimeter of the original rectangle is 120 meters. He also knows that the perimeter of the reduced rectangl

Whitepunk [10]

Answer:

24 meters is the width of the original rectangle.

Step-by-step explanation:

Given:

Bobby knows that the perimeter of the original rectangle is 120 meters. He also knows that the perimeter of the reduced rectangle is 30 meters and the reduced rectangle has a length of 9 meters.

Now, to get the width of original rectangle.

The reduced rectangle's perimeter = 30 meters.

The reduced rectangle's length = 9 meters.

Now, we find the width of reduced rectangle by using formula:

Let the width of reduced rectangle be $x.$