All categories

3 years ago

Identify the explicit function for the sequence in the table.

Mathematics

1 answer:

lord [1]3 years ago

3 0

Answer:

$a_n=9+(n-1)\cdot 5$ .

Step-by-step explanation:

The common difference is 5. The y values are going up by 5. So this is an arithmetic sequence since we have a common difference.

The explicit form for arithmetic sequence is:

$a_n=a_1+(n-1) \cdot d$ where d represents the commom difference and $a_1$ is the first term.

Here the first term is $a_1=9$ and we already determined the value for d which is 5.

Inputing these values for first term and common difference give:

$a_n=9+(n-1)\cdot 5$ .

You might be interested in

Combine like terms to create an equivalent expression -4/7p+(-2/7p)+1/7

STatiana [176]

-6/7p+1/7 um Idk what you wanted me to do but I added -4/7p with -2/7p

7 0

4 years ago

Read 2 more answers

Samantha walks 15 meters in 5 seconds. What is her walking rate? *<br> 1 point

Scrat [10]

Answer:

3 meters per second

Step-by-step explanation:

15 divided by 5 = 3

3 m/s

6 0

3 years ago

Read 2 more answers

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

4 years ago

Witch method will not get you to the point (5,2.5) ?

ICE Princess25 [194]

Answer:

that do it make any sense nshshshshshsjshshsnshdjdjjsjskslqpsjdbjdbsjsbsjdhshdhdhshsjsjsjjsjshshshsjs

Step-by-step explanation:

luttdlyjr5ie5uwriyrqkapeyEltezeltslrysrpfhzfhzllzryLRyLryYRh$&ffjzltjdltustudltustzfjxfj

5 0

4 years ago

Find the inverse function y=-5x+20

Nezavi [6.7K]

Answer:

$y= \frac{-1}{5} x +4$

Step-by-step explanation:

We can find the inverse function of this excercise, isolating X and then changing the variables.

$y=-5x+20\\y-20=-5x\\\frac{y-20}{-5} =x\\$

$\frac{y}{-5}-\frac{20}{-5} = x\\\frac{-1}{5}y+\frac{20}{5} =x\\ \frac{-1}{5}y+4 =x$

And now we change the variables and we get the inverse function:

$y= \frac{-1}{5} x +4$

7 0

3 years ago