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mel-nik [20]
3 years ago
6

The side of a square is Two raised to the seven halves power inches. Using the area formula A = s2, determine the area of the sq

uare.
Mathematics
2 answers:
AlladinOne [14]3 years ago
4 0

Answer:

Step-by-step explanation:

Area of a square = s²

s is the side length of the square

Given

s = 2^{7 1/2}

s = 2^15/2

Area = ( 2^15/2)²

Area = 2^15

Hence two area of the square is 2^15 inches

ololo11 [35]3 years ago
4 0

Answer:

A = 128 square inches

Step-by-step explanation:

A square is a figure with equal length of sides.

Given a square of side, s, 2^{\frac{7}{2} } inches.

Area of a square, A = length of side x length of side

                            = s x s

                            = s^{2}

Thus,

A = s^{2}

   = (2^{\frac{7}{2} } )^{2}

   = 2^{\frac{14}{2} }

   = 2^{7}

   = 128

A = 128 square inches

The are of the square is 128 square inches.

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8 0
3 years ago
Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke
Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that \sqrt{3}=\frac{m}{n} where  m and n are non negative integers, and the fraction \frac{m}{n} is irreducible, i.e., the numbers m and n have no common factors.

Now, squaring the equality at the beginning we get that

3=\frac{m^2}{n^2} (1)

which is equivalent to 3n^2=m^2. From this we can deduce that 3 divides the number m^2, and necessarily 3 must divide m. Thus, m=3p, where p is a non negative integer.

Substituting m=3p into (1), we get

3= \frac{9p^2}{n^2}

which is equivalent to

n^2=3p^2.

Thus, 3 divides n^2 and necessarily 3 must divide n. Hence, n=3q where q is a non negative integer.

Notice that

\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}.

The above equality means that the fraction \frac{m}{n} is reducible, what contradicts our initial assumption. So, \sqrt{3} is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, r\in\mathbb{Q}, which is equivalent to say that r=\frac{m}{n} where  m and n are non negative integers. Also, assume that k\in\mathbb{Z}. So, we want to prove that k\cdot r\in\mathbb{Z}. Recall that an integer k can be written as

k=\frac{k}{1}.

Then,

k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}.

Notice that the product mk is an integer. Thus, the fraction \frac{mk}{n} is a rational number. Therefore, k\cdot r\in\mathbb{Q}.

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have x is irrational and p\in\mathbb{Q}.

Write q=x+p and let us suppose that q is a rational number. So, we get that

x=q-p.

But the subtraction or addition of two rational numbers is rational too. Then, the number x must be rational too, which is a clear contradiction with our hypothesis. Therefore, x+p is irrational.

7 0
3 years ago
If y varies inversely as x, and y = 23 when x<br> 8, find y when x 4.
Virty [35]

Answer:

y = 46

Step-by-step explanation:

Given y varies inversely as x then the equation relating them is

y = \frac{k}{x} ← k is the constant of variation

To find k use the condition y = 23 when x = 8 , then

23 = \frac{k}{8} ( multiply both sides by 8 )

184 = k

y = \frac{184}{x} ← equation of variation

When x = 4 , then

y = \frac{184}{4} = 46

7 0
2 years ago
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Ymorist [56]

Answer:

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Step-by-step explanation:

5 0
3 years ago
I know y= 40 but what is x ?
puteri [66]

Answer:

x = 67; y = 40

Step-by-step explanation:

You are correct with y. Angles y + 23 and 63 are vertical, so they are congruent. That gives you y + 23 = 63, which gives y = 40.

To find x, look at angles 2x - 17 and 63.

They are adjacent angles whose outside rays lie on a line. Angles like these are called a linear pair, and they are supplementary. That means their measures add up to 180 deg.

2x - 17 + 63 = 180

2x + 46 = 180

2x = 134

x = 67

6 0
3 years ago
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