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2 years ago

You've won a new prize! Go to your game board.

Mathematics

2 answers:

erastovalidia [21]2 years ago

5 0

Answer:

quadrant l

Step-by-step explanation:

umka2103 [35]2 years ago

4 0

The point is located in quadrant I

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tipo de lote? para qué tema

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2 years ago

The amount of corn chips in a 16-ounce bag by the dispensing machine has a normal distribution with a mean of 16.5 ounces and a

Elenna [48]

Answer: 1+1=2

Step-by-step explanation: Imagine you have 1 cookie, and a friend gives you 1 more, that means your total amount of cookies is 2. So that proves that 1+1=2!

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2 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

Combine like terms: 3p2q2-3p2q3+4p2q3-3p2q2+pq PLEASE HELP!!! ASAP!!!

ch4aika [34]

Answer:

p²q³ + pq and pq(pq² + 1)

Step-by-step explanation:

Given

3p²q² - 3p²q³ +4p²q³ -3p²q² + pq

Required

Collect like terms

We start by rewriting the expression

3p²q² - 3p²q³ +4p²q³ -3p²q² + pq

Collect like terms

3p²q² -3p²q² - 3p²q³ +4p²q³ + pq

Group like terms

(3p²q² -3p²q²) - (3p²q³ - 4p²q³ ) + pq

Perform arithmetic operations on like terms

(0) - (-p²q³) + pq

- (-p²q³) + pq

Open bracket

p²q³ + pq

The answer can be further simplified

Factorize p²q³ + pq

pq(pq² + 1)

Hence, 3p²q² - 3p²q³ +4p²q³ -3p²q² + pq is equivalent to p²q³ + pq and pq(pq² + 1)

7 0

3 years ago

Do u agree or disagree and why?

pshichka [43]

Answer:

Erm. i would disagree

Step-by-step explanation:

Because they don't look the same or anything like that.

4 0

3 years ago

Read 2 more answers