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kondor19780726 [428]
2 years ago
9

ANSWER ASAP PLEASE!

Mathematics
2 answers:
aivan3 [116]2 years ago
6 0

Answer:

c. 2:7

Step-by-step explanation:

4+10 is 14

2 is the least common factor

4/2 simplified to 2

14/2 simplified to 7

zimovet [89]2 years ago
5 0

Answer:

2:7

Step-by-step explanation:

4+10=14

The least common factor is 2.

4/2=2

14/2=7

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Prove that 1/sin^2A -1/tan^2A= 1
Vika [28.1K]

Answer:

Step-by-step explanation:

LHS =\dfrac{1}{Sin^{2} \ A }-\dfrac{1}{Tan^{2} \ A }\\\\\\ = \dfrac{1}{sin^{2} \ A}- \dfrac{1}{\dfrac{Sin^{2} \ A}{Cos^{2} \ A}}\\\\\\= \dfrac{1}{sin^{2} \ A } - \dfrac{Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{1-Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{Sin^{2} \ A}{Sin^{2} \ A}\\\\\\= 1 = \ RHS

Hint: 1 - Cos² A = Sin² A

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2 years ago
The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV
JulijaS [17]

Answer:

\frac{dT}{dt}=3.78^{\circ}K/min

Step-by-step explanation:

We have to calculate the time derivative of T=PV/nR with P and V variable and n and R constants. This is:

\frac{dT}{dt} =\frac{d\frac{PV}{nR}}{dt}=\frac{1}{nR}\frac{d(PV)}{dt}

What we have to do is the derivative of a product:

\frac{d(PV)}{dt}=P\frac{dV}{dt}+V\frac{dP}{dt}

Substituting, we have:

\frac{dT}{dt} =\frac{P\frac{dV}{dt}+V\frac{dP}{dt}}{nR}

where all these values are given since the time derivatives of P and V are their variation rate, using minutes.

We then substitute everything, noticing that already everything is in the same system of units so they cancel out:

\frac{dT}{dt}=\frac{P\frac{dV}{dt}+V\frac{dP}{dt}}{nR}=\frac{(8atm)(0.16L/min)+(13L)(0.14atm/min)}{(10mol)(0.0821Latm/mol^{\circ}K)}

And then just calculate:

\frac{dT}{dt}=3.78^{\circ}K/min

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