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What value should go in the empty boxes to complete the calculation for finding the product of 0.61 × 0.45? Both numbers are the

Mekhanik [1.2K]

0.2745 I believe... because 0.61 times 0.45 gives you that answer....

6 0

3 years ago

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Find the area and the circumference of a circle with radius 9 ft.

Nitella [24]

Answers:

Area = 254.34 square feet

Circumference = 56.52 feet

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Work Shown:

Area = pi*(radius)^2

A = pi*r^2

A = 3.14*9^2

A = 254.34 square feet

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Circumference = 2*pi*(radius)

C = 2*pi*r

C = 2*3.14*9

C = 56.52 feet

The two results are approximate since pi = 3.14 is approximate.

5 0

3 years ago

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If a day was a year, how many does it take to make a million years?

Tasya [4]

It would take 2,740 days to make a million years.

1,000,000 divided by 365.

6 0

3 years ago

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Over the past several months, the water level of a lake has been decreasing by 3% each week. If the highest water level before t

VladimirAG [237]

Answer:

y= 407.54654 ft

Step-by-step explanation:

This is an exponential decay functions

y = ab^x where a is the initial value and b is ( 1 - rate of decay)

a = 520

b = 1- 3%

y = 520 ( 1- .3) ^x

y = 520 ( .97) ^x

We want x to be 8 weeks

y = 520 ( .97) ^8

y= 407.54654 ft

5 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

4 years ago