Ask question

Ask question

All categories

3 years ago

12

Ali was swimming along the bottom of his backyard pool at a depth of 6 feet then he pushed off the bottom and rose 4 feet what i

s ali's position now relative to the surface of the pool

1 answer:

olga nikolaevna [1]3 years ago

7 0

Answer:

2 feets

Step-by-step explanation:

Given that :

Alli's Initial depth = 6 feets

Number of feets in which Alli rose = 4 feets

Alli's new position relative to the surface of the pool:

Difference of Alli's initial depth and feets risen :

6 feets - 4 feets = 2 feets

Hence, Alli is now 2 feets below the surface of the pool.

You might be interested in

There are 20 baseball cards and 10 football cards on display at a sports museum. Noah says that

erica [24]

Answer:

They are both right becuase if you just take away 1 from 20 and 10 you get 2 and 1. For the 4:2 one take away 1 and mutiply by 2 equalling 4:2 but they are still the same thing.

Step-by-step explanation:

6 0

2 years ago

Whats 10 times 5 in the united states of america

OverLord2011 [107]

Answer:

50

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Calculate exactly how many miles not light year is the star Sirius is from the sun

algol13

Answer:

5.06208e+13 Hope it helps :)

Step-by-step explanation:

8 0

3 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

3 years ago

Mr. Toshiro manages a company that supplies a variety of domestic and imported nuts to supermarkets. He received an order for 12

alexandr1967 [171]

Answer:

n = 120i + 310j + 60k

p = 29i + 18j + 21 k

Total cost = $10,320

Step-by-step explanation:

Let n and p represent the vectors for number of products and prices respectively.

Also, the coordinates i,j and k represent cashews, walnut and Brazil nut respectively.

The vector form of the total number of bags ordered and the cost are;

n = 120i + 310j + 60k

p = 29i + 18j + 21 k

We can obtain the total cost by obtaining the dot product of the two vectors.

Total cost = n.p = (120i + 310j + 60k).(29i + 18j + 21 k)

C = 120×29 + 310×18 + 60×21

C = $10,320

7 0

3 years ago