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balu736 [363]
3 years ago
9

If you do this you are officially skilled: (SAT Prep) In the figure, if PN = LN, NP Is parallel to MQ, and QL bisects ∠PQM, what

is value of x?

Mathematics
2 answers:
kumpel [21]3 years ago
8 0

Answer:

67º

Step-by-step explanation:

IgorC [24]3 years ago
5 0

Answer:

Try the suggested solution, shown on the picture attached

Step-by-step explanation:

Note, 'm(MNO)' means m(∠MNO), and for issues 2, 4 ,5 - the described angles are angles inside the declared triangle.

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Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240<br><br> dx/dt=
katrin2010 [14]

Answer:

The value of \frac{dx}{dt} is \frac{160}{x^3}.

Step-by-step explanation:

The given equation is

x^4=8y^5-240

We need to find the value of \frac{dx}{dt}.

Differentiate with respect to t.

4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0              [\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]

4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

4x^3\frac{dx}{dt}=40(2)^4(1)

4x^3\frac{dx}{dt}=40(16)(1)

4x^3\frac{dx}{dt}=640

Divide both sides by 4x³.

\frac{dx}{dt}=\frac{640}{4x^3}

\frac{dx}{dt}=\frac{160}{x^3}

Therefore the value of \frac{dx}{dt} is \frac{160}{x^3}.

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Karen paid 43.70 for a book. The price of the books was $40. What was the sales tax rate?
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