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sashaice [31]
2 years ago
10

A sequence has a common ratio of Three-halves and f(5) = 81. Which explicit formula represents the sequence? f(x) = 24(Three-hal

ves) Superscript x minus 1 f(x) = 16(Three-halves) Superscript x minus 1 f(x) = 24(Three-halves) Superscript x f(x) = 16(Three-halves) Superscript x
Mathematics
2 answers:
mezya [45]2 years ago
9 0

Answer:

B

The other answer was too long.

Ede4ka [16]2 years ago
6 0

Answer:

f(x)=16\, (\frac{3}{2} )^{x-1}

Step-by-step explanation:

Since we are given the common ratio (3/2), all we need to find to define the geometric sequence, is its multiplicative factor (a) that corresponds to the first term of the sequence - remember that all consecutive terms will be generated by multiplying this first value repeatedly by the common ratio (3/2) as shown below:  

f(1)=a\\f(2)=a\,*\,(\frac{3}{2} )\\f(3)=a\,*\,(\frac{3}{2} )^2\\f(4)=a\,*\,(\frac{3}{2} )^3\\f(5)=a\,*\,(\frac{3}{2} )^4

Since we are given the information that f(5)=81 we can use this to find the value of the first term:

f(5)=\,a\,*\,(\frac{3}{2} )^4\\81=\,a\,*(\frac{81}{16} )\\a\,*\,81=\,81\.*\,16\\a\,=\,16

Notice as well that the first term doesn't contain the common ratio, the second term contains the common ration (3/2) to the power one, the third one contains the common ratio to the power two, the fourth one contains it to the power three, and so forth. So the exponent at which the common ratio appears is always one unit less than the order (x) of the term in question. This concept helps us finalize the expression for the sequence's formula:

f(x)=16\,*\,(\frac{3}{2} )^{x-1}

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Answer:

The height h is 5\ cm

Step-by-step explanation:

we know that

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SA=2B+Ph\\

where

B is the area of the base

P is the perimeter of the base

h is the height

In this problem

If fifty percent of the surface area of the bread is crust

then

2B=Ph\\

substitute the values

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5 0
2 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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A set of notations (SSS, SAS, ASA and RHS) is used to describe/prove that two triangles are conjurent. For each pair of triangle
lana66690 [7]

Answer:

The answer is given below

Step-by-step explanation:

The statements are not given but from this definition, you should be able to select the correct statement.

Triangles are said to be congruent if they have the same size and shape. A triangle is made up of six (three sides and three angles) but only three is usually needed to prove triangle congruence. These are:

a) Side - Side - Side (SSS): Two triangles are congruent if the length of their three sides are equal to each other.

b) Side- angle- side (SAS): If two sides and one angle of a triangle is equal to two sides and a angle of another triangle then the two triangles are congruent to each other.

c) Angle - side - angle (ASA): If two angles and a side of a triangle are equal to a corresponding two angles and a side of another triangle then the two triangles are congruenr.

d) Right angle - hypotenuse - side (RHS): If for two right angled triangles, their hypotenuse and a side (leg) are equal, then the two triangles are congruent.

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Cerrena [4.2K]

Answer:

4 years

Step-by-step explanation:

FYI this sounds like a personal problem

6 0
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Triangular numbers can be represented with equilateral triangles formed by dots. The first five triangular numbers are 1, 3, 6,
Alisiya [41]
Direct variation is y = kx  where y  would be the value of the number and x would be its position in the sequence, k is a constant 

so for consecutive values y/x would be a constant k 
In this case its not true becuse for example  3/1 = 3 and 6/3 = 2 

so there is no direct variation here.
7 0
2 years ago
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