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3 years ago

7

A cylinder has a height of h inches and a diameter of 3 inches. Give the expression of the volume of the cylinder in cubic inche

s.

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

Expression of volume = 1/3 × pi × radius
(squared ) × height

radius = diameter / 2

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Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|

Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0

3 years ago

You can use the formula c= 5/9 (f-32) to convert temperature in degrees fahrenheit, f, to temperature in degrees celsius,

mrs_skeptik [129]

C=5/9 (62-32)
c=(5/9)*30
c= 150/9
c=16.666....
so c=16.6

6 0

3 years ago

Read 2 more answers

Consider the function g(x) = 10/x

ss7ja [257]

<span><span>The correct answers are:
</span><span>(1) The vertical asymptote is x = 0
(2) The horizontal asymptote is y = 0

</span><span>Explanation:
</span><span>(1) To find the vertical asymptote, put the denominator of the rational function equals to zero.

Rational Function = g(x) = </span></span> $\frac{10}{x}$ <span>

Denominator = x = 0

Hence the vertical asymptote is x = 0.

(2) To find the horizontal asymptote, check the power of x in numerator against the power of x in denominator as follows:

Given function = g(x) = </span> $\frac{10}{x}$ <span>

We can write it as:

g(x) = </span> $\frac{10 * x^0}{x^1}$ <span>

If power of x in numerator is less than the power of x in denomenator, then the horizontal asymptote will be y=0.
If power of x in numerator is equal to the power of x in denomenator, then the horizontal asymptote will be y=(co-efficient in numerator)/(co-efficient in denomenator).
If power of x in numerator is greater than the power of x in denomenator, then there will be no horizontal asymptote.

In above case, 0 < 1, therefore, the horizontal asymptote is y = 0
</span>

5 0

4 years ago

Read 2 more answers

Sin(x+y)-sin(x-y)/cos(x+y)+cos(x-y)=tany<br> how do i verify this?

dsp73

Answer:

Step-by-step explanation:

Numerator

sin

x

cos

y

+

cos

x

sin

y

−

[

sin

x

cos

y

−

cos

x

sin

y

)

=

sin

x

cos

y

+

cos

x

sin

y

−

sin

x

cos

y

+

cos

x

sin

y

=

2

cos

x

sin

y

Denominator

cos

x

cos

y

−

sin

x

sin

y

+

cos

x

cos

y

+

sin

x

sin

y

=

cos

x

cos

y

−

sin

x

sin

y

+

cos

x

cos

y

+

sin

x

sin

y

=

2

cos

x

cos

y

---------------------------------------------------------------

left side can now be expressed as

2

cos

x

sin

y

2

cos

x

cos

y

=

2

cos

x

sin

y

2

cos

x

cos

y

=

sin

y

cos

y

and

sin

y

cos

y

=

tan

y

=

right side hence proved

7 0

3 years ago

What is the interquartile range of this data? 6 8 9

vladimir1956 [14]

Answer:

24

Step-by-step explanation:

Just subtract the first quartile form the third quartile.

third quartile - first quartile

32 - 8 = 24 IQR

FIRST quartile: 8

SECOND quartile: 20

THIRD quartile: 32

So the IQR is 24

Hope this helps :)

3 0

3 years ago

Read 2 more answers