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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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Darya [45]

Answer: 6/5

Step-by-step explanation:

7 0
2 years ago
i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

3 0
3 years ago
Which factors of -12 add to -1?
beks73 [17]

Answer:

-4 and 3

Step-by-step explanation:

4 0
3 years ago
Finding the values of Product and Quotient Functions:
Goshia [24]

Answer:

2

Step-by-step explanation:

r(x) = 2 sqrt(x)

s(x) = sqrt(x)

r(x) / s(x) =  2 sqrt(x) / sqrt(x)

              = 2

8 0
3 years ago
Given that P(A)=0.5 and p(AUB)=0.6 find the P(B) if P(A/B) =0.4?
Yuki888 [10]
First we have to find p(AnB)
P(A/B)=p(AnB)/p(A)
P(AnB)=p(A) x p(A/B)
P(AnB)=0.5 x 0.4 = 0.2
Then to find p(B)
P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = 0.5+p(B)-0.2
0.6=0.3+p(B)
P(B)=0.6-0.3=0.3

4 0
2 years ago
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