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3 years ago

The first three terms of a sequence are given. Round to the nearest thousandth (if

Mathematics

1 answer:

Fofino [41]3 years ago

6 0

Answer:

19683

Step-by-step explanation:

1. find the ratio

this is geometric so..

54÷18=3

18÷6=3

2. set up the problem

3(3)^9-1

3^8= 6561

6561(3)=19683

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Which expression is equal to<br> -18b^-4

victus00 [196]

Answer:

$- \frac{18}{ {b}^{4} }$

Step-by-step explanation:

$- 18b {}^{ - 4}$

express b^-4 with a positive exponent using:

$a {}^{ - n} = \frac{1}{ {a}^{n} }$

$- 18 \frac{1}{ {b}^{4} }$

$- \frac{18}{ {b}^{4} }$

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3 years ago

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PLS HELP the LAST QUESTION, please...

Alex73 [517]

Answer:

280 Cubic inches

Step-by-step explanation:

When finding the volume, the equation is V=l*w*h

Therefore, the equation should be:

10.1*3.9*6.8

This is equal to exactly 267.852

However, the answer is 280 cubic in. because it's a larger number and it's about 13 from the actual answer. On the other hand, 250 cubic in. is about 17 from the actual number.

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3 years ago

How does the graph of g(x) = 3x – 2 compare to the graph of f(x) = 3x?

Novosadov [1.4K]

Step-by-step explanation:

It is shifted down 2 units.

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3 years ago

Read 2 more answers

Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap

pantera1 [17]

Answer:

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e $\dfrac{dV}{dt}= 20 \ ft^3/min$

we know that radius r is always twice the diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

$V = \dfrac{\pi r^2 h}{3}$

$V = \dfrac{\pi (h/2)^2 h}{3}$

$V = \dfrac{1}{12} \pi h^3$

$V = \dfrac{ \pi h^3}{12}$

Taking the differentiation of volume V with respect to time t; we have:

$\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}$

$\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}$

we know that:

$\dfrac{dV}{dt}= 20 \ ft^3/min$

So;we have:

$20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}$

$20= 56.25 \pi \times \dfrac{dh}{dt}$

$\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

8 0

3 years ago