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3 years ago

3. If the line tangent to the graph of the function f at the point (1,7) also passes through the point

Mathematics

1 answer:

Usimov [2.4K]3 years ago

5 0

Answer:

f'(-1)= 3

Step-by-step explanation:

slope = (y2 - y1)/(x2 - x1)= (7-(-2))/(1-(-2))=9/3=3

Slope of the tangent line in the given point equals f' at this point.

So,

f'(-1)= 3

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Verify the identity:<br><br> cos(a+B)/cos a sin B = cot B -tan a

alexgriva [62]

Cos( A + B ) = cosAcosB - sinAsinB ;
cos( A + B ) / ( cosAsinB ) = ( cosAcosB - sinAsinB ) / ( cosAsinB ) = ( cosAcosB ) / ( cosAsinB ) - ( sinAsinB ) / ( cosAsinB ) = cosB / sinB - sinA / cosA = cotB - tanA ;

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3 years ago

After heating up in a teapot, a cup of hot water is poured at a temperature of 204°F.

laila [671]

Answer:

Question 1: 0.081 min⁻¹

Question 2: 168ºF

Explanation:

The missing equation is:

$T=T_a+(T_0-T_a)e^{-kt}$

You are given:

Ta = the temperature surrounding the object = 73ºF
T₀ = the initial temperature of the object = 204ºF
t = 2.5 min
T after 2.5 min = 180ºF

Question 1: Find the the value of k.

Substitute the data into the equation and solve for k:

$180\º=73\º+(204\º-73\º)e^{-k\cdot 2.5min}\\ \\ 107\º=131\ºe^{-k\cdot 2.5min}\\\\-k\cdot 2.5min=\ln (107/131)\\\\k=0.081min^{-1}$

Question 2: Find the Fahrenheit temperature of the cup of water, to the nearest degree, after 4 minutes.

Subsititute into the formula and compute:

$T=73\ºF+(204\ºF-73\ºF)e^{-0.081min^{-1}\times 4min}\\ \\ T=73\ºF+(131\ºF)(0.72325024)\\\\T=168\ºF$

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3 years ago

1/3(x+2/3) =1/6(x-4) thanks

zheka24 [161]

1/3+2/9=1/6x-4/6
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Answer:

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Step-by-step explanation:

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3 years ago

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Point a is located in Quadrant 1

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3 years ago

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