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3 years ago

The limit represents the derivative of some function f at some number a. State such an f and a.

1 answer:

7 0

Limit definition of the Derivative of a function: $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ At x = a: $\lim_{h \to0} \frac{f(a+h)-f(a)}{h}$
If: h = x -π/4, we will have: $\lim_{h \to \pi /4} \frac{tan ( \pi /4 +h)-1}{h}$ When we compare it with a limit definition:
f( x )= tan x
f( a ) = 1
tan ( a ) = 1, a = π/4

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H E L P!!! PLEASE HELP ME!!! HURRRRRRY!!!!

borishaifa [10]

Answer:

h= 17/2

Step-by-step explanation:

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3 years ago

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So far, Phillip has completed 3/4 of his run. If he has run 9 miles, how far does he plan to run

uysha [10]

Answer:

He has to run a total of 12 miles and has 3 more miles to run

Step-by-step explanation:

Well we know that 3/4 is 9 out of the total miles he is willing to run. We can divide 9 by 3 to get 1

Now we know 1/4 of his run is 3 miles

3*4= 12

He has to run a total of 12 miles and has 3 more miles to run

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Find 437 th term of the arithmetic sequence whose first term is

Brrunno [24]

The 437th term of arithmetic sequence is 36.

Step-by-step explanation:

First term = $a_{1}$ = -73

Common difference = d = $\frac{1}{4}$

Term to find = 437th

Therefore,

n= 437

The formula for arithmetic sequence is;

$a_{n}=a_{1}+(n-1)d\\$

Putting values;

$a_{437}=-73+(437-1)(\frac{1}{4})\\a_{437}=-73+(436)(\frac{1}{4}\\a_{437}=-73+\frac{436}{4}\\a_{437}=-73+109\\a_{437}=36$

The 437th term of arithmetic sequence is 36.

Keywords: Arithmetic sequence, common difference

Learn more about arithmetic sequences at:

#LearnwithBrainly

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3 years ago

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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2 years ago

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Mrs. Martinez teaches Spanish part time at a nearby college. In one class, there are 12 women and 14 men; in her other class, th

AleksandrR [38]

Answer: Women =23. Men =31. All together equals =54

Step-by-step explanation:

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