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3 years ago

What is the next term in the geometric sequence? 54,36,24

Mathematics

2 answers:

katovenus [111]3 years ago

8 0

I think 16...............

vodomira [7]3 years ago

7 0

Answer: 16

Step-by-step explanation:

The formular for calculating the nth term of a geometric progression is arⁿ⁻¹

WHERE a= first term, r = common ratio and n= number of term

a = 54

since we are looking for the fourth term n = 4

To get our common ratio; we simply divide a number by its preceding number

r = 36/54 = 2/3

our fourth term is;

Tn = arⁿ⁻¹

T₄ = 54 (2/3)⁴⁻¹

= 54 (2/3)³

=54(8/27)

= 432/27

=16

Therefore the next term is 16

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3 years ago

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3 years ago

56 = h/9<br> k/5 - 10 = 3<br> 3t + 5 =2

uranmaximum [27]

Answer:

h = 504, k = 65, t = -1

Step-by-step explanation:

56 = h/9

h = 56 x 9

h = 504

k/5 - 10 = 3

k/5 = 3 + 10

k/5 = 13

k = 13 x 5

k = 65

3t + 5 = 2

3t = 2 - 5

3t = -3

t = -3/3

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2 years ago

A fence for a rectangular garden with one side against an existing wall is constructed by using 60 feet of fencing. What is the

Y_Kistochka [10]

Step-by-step explanation:

A fence for a rectangular garden with one side against an existing wall is constructed by using 60 feet of fencing.

Perimeter of rectangle (3 sides)= 60 feet

Let 'x' be the width of the wall

$Perimeter = 2(length)+2(width)\\60=2(length)+2x\\\\60-2x=2(length)\\\frac{60-2x}{2} =length\\Length =30-x$

Formula for the area of the rectangle is

$Area=length \cdot width\\A=length(x)$

Replace the length we got using perimeter

$A=(30-x)(x)\\A(x)= 30x-x^2$

To find out the maximum are we take derivative

$A'(x)= 30-2x\\0=30-2x\\-30=-2x\\x=15$

find out second derivative to check whether x=15 is maximum

$A''(x)=-2$

second derivative is negative

So, Maximum area at x=15

To find maximum area we plug in 15 for x in A(x)

$A(x)=30x-x^2\\A(15)=30(15)-15^2\\A(15)=225$

So, maximum area is 225 square feet

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3 years ago