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2 years ago

Please help me! thank you :)

Mathematics

1 answer:

Papessa [141]2 years ago

6 0

Answer:

Step-by-step explanation:

As the equation has already been factorised, we can find the value of x that results in 0 in each bracket:

x - 3 >= 0

x >= 3

x+2 >= 0

x >= -2

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Which quadrilaterals are also parallelograms?

sveticcg [70]

A, B, and D. A trapezoid cannot be parallel because the 2 left and right sides will eventually meet/connect.

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2 years ago

large cheese pizzas cost $5 each and large one topping pizzas cost $6 each. write an equation that represents the total cost, T

Gelneren [198K]

Answer:

T = 5c +5d, where unit of the equation is $.

Step-by-step explanation:

Cost of 1 large cheese pizza = $5

No. of large cheese pizza = c

therefore

cost of c large cheese pizza = c*Cost of 1 large cheese pizza = $5c

Cost of 1 large one topping pizza = $6

No. of large one topping pizza = d

therefore

cost of d large one topping pizza = d*Cost of 1 large one topping pizza = cost of d large one topping pizza = $6d

Total cost of c large cheese pizza and cost of d large one topping pizza

= $5c +$6d

total cost is represented by T

thus, T = $5c +$5d Answer

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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 $\frac{dP}{dt}$

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