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3 years ago

6

Find the 5th term of the sequence in which T1=8 and Tn=3t n-1

1 answer:

yan [13]3 years ago

6 0

T_n = 3 * T_(n-1)

Long way (always works!)
T_5 = 3*T_4,
T_4 = 3*T_3
T_3 = 3*T_2
T_2 = 3*T_1

T_5 = 3*3*3*3*T_1 = 81*T_1 = 81*8 = 648!

Short way (sometimes it works!)

T_n = 3^(n-1) * T_1 (this case is a geometric series of ratio-=3)
T_5 = 3^4*8 = 648

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Answer:

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Step-by-step explanation:

Consider the provided monomial.

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3. A balancing balloon toy is in the shape of a hemisphere (half-sphere) attached to the base of a cone. If the toy is 4ft tall

Katen [24]

Answer:

The volume of the toy is $V=5.23\ ft^3$

Step-by-step explanation:

step 1

Find the volume of the hemisphere

The volume of the hemisphere is given by the formula

$V=\frac{2}{3}\pi r^{3}$

In this problem, the wide of the toy is equal to the diameter of the hemisphere

so

$D=2\ ft$

$r=2/2=1\ ft$ ----> the radius is half the diameter

substitute

$V=\frac{2}{3} \pi (1)^{3}=\frac{2}{3} \pi\ ft^3$

step 2

Find the volume of the cone

The volume of the cone is given by

$V=\frac{1}{3}\pi r^{2}h$

we know that

The radius of the cone is the same that the radius of the hemisphere

so

$r=1\ ft$

The height of the cone is equal to subtract the radius of the hemisphere from the height of the toy

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substitute the given values

$V=\frac{1}{3}\pi (1)^{2}(3)=\pi\ ft^3$

step 3

Find the volume of the toy

we know that

The volume of the toy, is equal to the volume of the cone plus the volume of the hemisphere.

so

$V=(\frac{2}{3} \pi+\pi)\ ft^3$

$V=(\frac{5}{3}\pi)\ ft^3$

assume

$\pi=3.14$

$V=\frac{5}{3}(3.14)=5.23\ ft^3$

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4 years ago

I need help please with this 1 problem

Lyrx [107]

Answer:

See below, please

Step-by-step explanation:

We know

$f(x) = {x}^{2} + 3x - 7$

$g(x) = 3x + 5$

$h(x) = 2 {x}^{2} - 4$

Hence

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$h(g(x))$

$= 2 \times {(3x + 5)}^{2} - 4$

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$= h(x) - f(x)$

$= (2 {x}^{2} - 4) - ( {x}^{2} + 3x - 7)$

$= {x}^{2} - 3x + 3$

$(f + g)(x)$

$= f(x) + g(x)$

$= ( {x}^{2} + 3x - 7) + (3x + 5)$

$= {x}^{2} + 6x - 2$

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