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3 years ago

Find the smallest perimeter and the dimensions for a rectangle with an area of 36 in^2

Mathematics

1 answer:

Tpy6a [65]3 years ago

3 0

This is a square with area 36 in^2

So dimensions are 6 * 6 ins and perimeter = 24 ins.

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The first term of a geometric sequence is 32, and the 5th term of the sequence is 818 .

sammy [17]

Answer:

$32,24,18,\frac{27}{2} ,\frac{81}{8}$

Step-by-step explanation:

Let x, y , and z be the numbers.

Then the geometric sequence is $32,x,y,z,\frac{81}{8}$

Recall that term of a geometric sequence are generally in the form:

$a,ar,ar^2,ar^3,ar^4$

This implies that:

a=32 and $ar^4=\frac{81}{8}$

Substitute a=32 and solve for r.

$32r^3=\frac{81}{8}$

$r^4=\frac{81}{256}$

Take the fourth root to get:

$r=\sqrt[4]{\frac{81}{256} }$

$r=\frac{3}{4}$

Therefore $x=32*\frac{3}{4} =24$

$y=24*\frac{3}{4} =18$

$z=18*\frac{3}{4} =\frac{27}{2}$

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