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3 years ago

HELP ME ASAP PLEASE, I DONT KNOW THE ANSWER.

Mathematics

1 answer:

Fudgin [204]3 years ago

7 0

64° i think because. its an iscoceles triangle so the bottom two angles are the same so you do 58+58=116 and because all angles in a triangle add to 180° you do 180-116=64°

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PLS HELP IF YOU ANSWER I WILL GIVE YOU BRAINLIEST ANSWER

Elodia [21]

Answer:

4.5

Step-by-step explanation:

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3 years ago

Is 6/7 greater than 4/6?

Oxana [17]

Yes, because 6/7 is the same as 36/42 and 4/6 is the same as 28/42. It is easy to see that 6/7 is greater than 4/6

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3 years ago

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Triangle XYZ has vertices X(1, 3), Y(0, 0), and Z(–1, 2). The image of triangle XYZ after a rotation has vertices

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4 years ago

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#2 Find the area.<br> 10ft

den301095 [7]

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It would be 78.539...

Step-by-step explanation:

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3 years ago

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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