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Georgia [21]
3 years ago
13

Visualize the following procedure: Tear in half a square piece of paper with an area of one square unit. Then tear it in half ag

ain. Predict the area of one of the resulting rectangles after 5 tears.
Mathematics
2 answers:
7nadin3 [17]3 years ago
8 0

Answer:

= 0.0625 in²

Step-by-step explanation:

Area of a piece of paper = 1 unit²

then we tear this paper into half.

So area becomes 0.5 unit²

In this way the sequence formed will be

1, 0.5, 0.25, .........

This sequence is a geometric sequence

as \frac{T_{2} }{T_{1}}= \frac{0.5}{1}=0.5

common ratio r = 0.5

We have to find the 5th term of this sequence

Explicit formula of a geometric sequence is represented by

             T_{n} =a(r)^{n-1}

Now      T_{5} =1(0.5)^{5-1}

                   = (0.5)⁴

                   = 0.0625 in²

After 5 tears area would be  0.0625 in²

Nadusha1986 [10]3 years ago
4 0

Answer:

Area = \frac{1}{16} square units

Step-by-step explanation:

To predict the area of the piece of paper after 5 tears we can use a geometric sequence.

Each time a piece is torn, half of the previous area is lost. If the area of the first piece is 1 square unit then:

a_1 = 1\\\\r = \frac{1}{2}\\\\a_2 = 1 *\frac{1}{2}\\\\a_2 = \frac{1}{2}\\\\a_n = 1(\frac{1}{2})^{n-1}\\\\a_5 = 1(\frac{1}{2})^{5-1}

a_5 = \frac{1}{16} square units

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Simplify:
Serga [27]

Answer:

(a) 16-2x

(b)-x-2 or x+6

(c)

Step-by-step explanation:

(a)

|6-x|+|x-10|,

|4-x|=4-x,so x<4

|2-x|=x-2=-(2-x)

2-x<0

2<x

or x>2

2<x<4

so |6-x|=6-x,in 2<x<4

|x-10|=-(x-10),in 2<x<4

|6-x|+|x-10|=6-x-(x-10)=6-x-x+10=16-2x

(b)

2-|2-|x-2||,if x<-2

|x-2|=-(x-2) if x<-2

=2-|2-{-(x-2)}|

=2-|2+x+2|

x<-2

x+2<0

=2-|x+4|

=2-(x+4),if x>-4,or -4<x<-2

=2-x-4

=-x-2

if x<-4

then |x+4|=-(x+4)

2-|x+4|=2-{-(x+4)}

=2+x+4

=x+6

3 0
3 years ago
List the theorems for finding zeros of higher degree polynomials
astra-53 [7]
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8 0
3 years ago
I need help with these two problems !!
wolverine [178]

Not so sure about the second one, but the first one is pretty simple:

The area of a circle = pi x radius^2

Since you need to find the area of the entire dvd, I'm assuming they mean if the hole in the center wasn't there, so add that to the diameter.

12+1.5 = 13.5

Half of the diameter is the radius, so half of 13.5 = 6.75

Now, multiply pi (3.14) by the radius, but have the radius squared. You'd write the equation like this:

3.14 x 6.75^2 (or on paper, the ^2 would be a small two diagonal from the 5)

Actual answer: 143.06625

You can round that to 143.

4 0
3 years ago
Can anyone help me choose the correct answer?!
lara [203]

Answer:

f(x)=x-6 is your answer.

when x=0

f(0)=0-6=-6

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7 0
3 years ago
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
3 years ago
Read 2 more answers
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