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3 years ago

What is the sequence in a pattern?

1 answer:

8 0

A sequence is a list of numbers, geometric shapes or other objects, that follow a specific pattern. The individual items in the sequence are called terms , and represented by variables like x n .

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The equation shows the relationship between x and y:

Masja [62]

I believe the answer is -5

5 0

3 years ago

6. How much heat is lost cooling 25 grams of water from 90°C to 45°C?

sergey [27]

Answer:

4,7025 (kJ).

Step-by-step explanation:

1) according to the condition: m=25 gr.; t₂=90; t₁=45; c=4.18 J/(g°C);

2) the formula of required heat is

Q=c*m*(t₂- t₁);

3) according to the formula of heat:

Q=4.18*25*45=4702.5 (J).

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3 years ago

Guys help me what is 137x8 multiply with expanded form???????????? Please help!!!!!!!!!!!!!!!!!! 33 points for the person who he

Helen [10]

Answer:

maybe 20

Step-by-step explanation:

or find it out yourself

4 0

3 years ago

Please answer this question (fill In the blanks)

alisha [4.7K]

1) Associative

2) Reversal

3) Addition

4) Commutative

5) Multiplicative Inverse

6) Additive Inverse

7) Reversal

8) Substitution

<u>RATE AS BRAINLIEST</u>

8 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