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2 years ago

Given the geometric sequence where a1 = 3 and r = √2 find a9

Mathematics

1 answer:

Zigmanuir [339]2 years ago

8 0

Answer:

$a_9=48$

Step-by-step explanation:

we are given

sequence is geometric

so, we can use nth term formula

$a_n=a_1(r)^{n-1}$

we have

$a_1=3$

$r=\sqrt{2}$

we have to find a9

so, we can plug n=9

we get

$a_9=3(\sqrt{2})^{9-1}$

$a_9=2^4\cdot \:3$

$a_9=48$

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Answer:

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2 years ago

Classify each graph as increasing, decreasing, constant

posledela

Answer:

3. increasing

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Step-by-step explanation:

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8 0

1 year ago

What is the simplified expression for -3(2x-y)+2y + 2(x+y)?

slamgirl [31]

Answer:

$- 3(2x - y) + 2y + 2(x + y) \\ - 6x + 3y + 2y + 2x + 2y \\ = - 4x + 7y$

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6 0

2 years ago

Read 2 more answers

Dr. Manion a mixed 9.357G of chemical A, 12.082g of chemical b,

Gnoma [55]

Amount of chemical A mixed in medicine = 9.357 g

Amount of chemical B mixed in medicine = 12.082 g

Amount of chemical C mixed in medicine = 7.502 g

We have to determine the total amount of medicine he prepared in grams.

Total amount of medicine

= 9.357 + 12.082 + 7.502

= 28.941 grams

So, he prepared 28.941 grams of medicine.

Now, we have to round each chemical to the nearest tenth.

Amount of chemical A mixed in medicine = 9.357 g

Rounding to nearest tenth = 9.4 grams

Amount of chemical B mixed in medicine = 12.082 g

Rounding to nearest tenth = 12.1 grams

Amount of chemical C mixed in medicine = 7.502 g

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6 0

3 years ago

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

3 years ago