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2 years ago

Someone please help me with 16-18 please it’s due in 30 mins

Mathematics

1 answer:

AlekseyPX2 years ago

3 0

9514 1404 393

Answer:

16. 134°

17. 20°

18. 94°

Step-by-step explanation:

An inscribed angle is half the measure of the arc it intercepts.

16. Arc QT is twice the measure of inscribed angle QST, given as 67°.

Arc QT = 2×67°

Arc QT = 134°

17. Angle STR is half the measure of arc SR, so is ...

m∠STR = (arc SR)/2 = 40°/2

m∠STR = 20°

18. Arc RST is a semicircle, so has a measure of 180°. Then arc ST is the difference between that and arc RS:

arc ST = 180° -40° = 140°

Angle RST subtends a semicircle, so is 180°/2 = 90°. We know angle QST is 67°, so the remaining portion, angle QSR must be the complement of that:

m∠QSR = 90° -67° = 23°

Arc RQ will be twice this measure, or 46°.

So, the desired difference is ...

mST -mRQ = 140° -46° = 94°

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NARA [144]

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Power, product and chain rules:

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Product and chain rules:

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$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

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Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

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$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

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$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

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