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2 years ago

12

If you on a motorcycle trip of 30 miles in the and 5\19 of the trip is downhill, how many miles of the trip are not downhill? Re

duce to the lowest term.

1 answer:

Bess [88]2 years ago

8 0

Answer:

1/4 of the trip is not down hill

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How do i show my work for 12.62 divide by 0.4

insens350 [35]

3.155

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3 0

2 years ago

Prove that if n is a perfect square then n + 2 is not a perfect square

notka56 [123]

Answer:

This statement can be proven by contradiction for $n \in \mathbb{N}$ (including the case where $n = 0$ .)

$\text{Let $n \in \mathbb{N}$ be a perfect square}$ .

$\textbf{Case 1.} ~ \text{n = 0}$ :

$\text{$n + 2 = 2$, which isn't a perfect square}$ .

$\text{Claim verified for $n = 0$}$ .

$\textbf{Case 2.} ~ \text{$n \in \mathbb{N}$ and $n \ne 0$. Hence $n \ge 1$}$ .

$\text{Assume that $n$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $a \in \mathbb{N}$ s.t. $a^2 = n$}$ .

$\text{Assume $\textit{by contradiction}$ that $(n + 2)$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $b \in \mathbb{N}$ s.t. $b^2 = n + 2$}$ .

$\text{$n + 2 > n > 0$ $\implies$ $b = \sqrt{n + 2} > \sqrt{n} = a$}$ .

$\text{$a,\, b \in \mathbb{N} \subset \mathbb{Z}$ $\implies b - a = b + (- a) \in \mathbb{Z}$}$ .

$\text{$b > a \implies b - a > 0$. Therefore, $b - a \ge 1$}$ .

$\text{$\implies b \ge a + 1$}$ .

$\text{$\implies n+ 2 = b^2 \ge (a + 1)^2= a^2 + 2\, a + 1 = n + 2\, a + 1$}$ .

$\text{$\iff 1 \ge 2\,a $}$ .

$\text{$\displaystyle \iff a \le \frac{1}{2}$}$ .

$\text{Contradiction (with the assumption that $a \ge 1$)}$ .

$\text{Hence the original claim is verified for $n \in \mathbb{N}\backslash\{0\}$}$ .

$\text{Hence the claim is true for all $n \in \mathbb{N}$}$ .

Step-by-step explanation:

Assume that the natural number $n \in \mathbb{N}$ is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number $a$ ( $a \in \mathbb{N}$ ) such that $a^2 = n$ .

Assume by contradiction that $n + 2$ is indeed a perfect square. Then there should exist another natural number $b \in \mathbb{N}$ such that $b^2 = (n + 2)$ .

Note, that since $(n + 2) > n \ge 0$ , $\sqrt{n + 2} > \sqrt{n}$ . Since $b = \sqrt{n + 2}$ while $a = \sqrt{n}$ , one can conclude that $b > a$ .

Keep in mind that both $a$ and $b$ are natural numbers. The minimum separation between two natural numbers is $1$ . In other words, if $b > a$ , then it must be true that $b \ge a + 1$ .

Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by $(a + 1)$ and use the fact that $b \ge a + 1$ to make the left-hand side $b^2$ .)

$b^2 \ge (a + 1)^2$ .

Expand the right-hand side using the binomial theorem:

$(a + 1)^2 = a^2 + 2\,a + 1$ .

$b^2 \ge a^2 + 2\,a + 1$ .

However, recall that it was assumed that $a^2 = n$ and $b^2 = n + 2$ . Therefore,

$\underbrace{b^2}_{=n + 2)} \ge \underbrace{a^2}_{=n} + 2\,a + 1$ .

$n + 2 \ge n + 2\, a + 1$ .

Subtract $n + 1$ from both sides of the inequality:

$1 \ge 2\, a$ .

$\displaystyle a \le \frac{1}{2} = 0.5$ .

Recall that $a$ was assumed to be a natural number. In other words, $a \ge 0$ and $a$ must be an integer. Hence, the only possible value of $a$ would be $0$ .

Since $a$ could be equal $0$ , there's not yet a valid contradiction. To produce the contradiction and complete the proof, it would be necessary to show that $a = 0$ just won't work as in the assumption.

If indeed $a = 0$ , then $n = a^2 = 0$ . $n + 2 = 2$ , which isn't a perfect square. That contradicts the assumption that if $n = 0$ is a perfect square, $n + 2 = 2$ would be a perfect square. Hence, by contradiction, one can conclude that

$\text{if $n$ is a perfect square, then $n + 2$ is not a perfect square.}$ .

Note that to produce a more well-rounded proof, it would likely be helpful to go back to the beginning of the proof, and show that $n \ne 0$ . Then one can assume without loss of generality that $n \ne 0$ . In that case, the fact that $\displaystyle a \le \frac{1}{2}$ is good enough to count as a contradiction.

7 0

2 years ago

How does a digit in the ten thousand place compare to a digit in the thousands place

Butoxors [25]

A digit in the ten thousand place has a value of 10,000 times the value of the mere digit. While a digit in the thousands place has a value 1,000 times the value of the digit. So to compare you can do 10,000 / 1,000 = 10, which means that<span> a digit in the ten thousand place values ten times what the same digit values is it is the thousand place. </span>

4 0

2 years ago

Solve for x. 16x - 5 = 18x - 2 and Solve for x.<br><br>13x + 7 = 5x - 20

Naily [24]

The answer to the question

7 0

2 years ago

A building 62 feet tall casts a shadow 21 meters long . At what angl is the sun shining on the building

yulyashka [42]

Answer:

71.3°

Step-by-step explanation:

A building 62 feet tall casts a shadow 21 meters long . At what angle is the sun shining on the building?

We solve this question using the Trigonometric function of Tangent

tan x = Opposite/Adjacent

Where

Opposite = 62ft

Adjacent = 21 ft

x = Angle

Hence,

tan x = 62/21

x = arc tan (62/21)

x = 71.288262125°

Approximately = 71.3°

6 0

2 years ago