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3 years ago

2 questions plz Both will be in the comments

Mathematics

1 answer:

lesantik [10]3 years ago

7 0

The two questions aren’t there?

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A)125<br> B)130<br> C)120<br> D)135

Kipish [7]

Answer:

The answer is B.

Step-by-step explanation:

Pentagons have a sum of 540°

540 - (115 + 115 + 90 + 90)

540 - 410

130

7 0

3 years ago

Read 2 more answers

I don’t understand need help

kodGreya [7K]

Answer:

6y^2-12y-4

Step-by-step explanation:

(8y^2-5y+7) - (2y^2+7y+11)

(8y^2-5y+7)-2y^2-7y-11

Combine like terms:

6y^2-12y-4

5 0

4 years ago

Read 2 more answers

Rewrite 3/35 and 1/10 so they have common denominators

Aleks [24]

Answer:

$\frac{6}{70}$ and $\frac{7}{70}$

Step-by-step explanation:

The LCM of 35 and 10 is 70.

35×2=70

10×7=70

You got to multiply 3 with 2 and 1 with 7 to make 6/70 and 7/70.

5 0

3 years ago

A home improvement store advertises 50 square feet of flooring for $300.00 including a $100.00 installation fee. How much does e

dimulka [17.4K]

$6 . It will cost $6 because spending 6 dollars , 50 times will equal $300

3 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

4 years ago

2 questions plz Both will be in the comments​

2 questions plz Both will be in the comments