Ask question

Ask question

All categories

3 years ago

5

Qual a altura máxima atingida por um projétil cuja trajetória pode ser descrita pela função: h(x) = –4x² + 5, sabendo que h é a

altura do projétil e que x é a distância percorrida por ele, em metros?
A))5
B))6
C))7
D))8
E))9

1 answer:

masya89 [10]3 years ago

6 0

Answer:

A altura máxima é 5

Step-by-step explanation:

Matematicamente, a altura máxima pode ser obtida

A altura máxima é simplesmente o vértice da parábola

Começamos diferenciando a função Isso será -8x

Agora, defina -8x como 0

Isso significa que x = 0

Agora, substitua x = 0 de volta na equação temos

f (0) = -4 (0) ^ 2 + 5

f (0) = 5

A altura máxima é 5

You might be interested in

-3(n+9)=15<br> i need the step of solving this plzz brainliest if right!!

raketka [301]

Answer:

n = -14

Step-by-step explanation:

-3(n+9)=15

Distribute -3 to (n+9):

-3n - 27 = 15

Add 27 to both sides:

-3n = 42

Divide both sides by -3:

n = -14

5 0

3 years ago

Help meeeee!!!!!!! I’ll give you a cookie.

lianna [129]

Answer:

(-5,24)

Step-by-step explanation:

2f+3g-h

first we substitute in the numbers

2(-3)+3(1)-2

-6+3-2

-3-2

-5

Then again

2(5)+3(4)-(-2)

10+12+2

22+2

24

hope this helps!:)

7 0

3 years ago

schepotkina [342]

Answer:

3.56 cm.

Step-by-step explanation:

1/3 π r^2 h = V

1/3 π (5.13)^2 h = 98

h = 98 / (1/3 π (5.13)^2)

= 98 / 27.55899

= 3.556.

3 0

3 years ago

Stacy started with 40 dollars in a savings account that earns 5% annually. How much interest will she get in one year

umka2103 [35]

Answer:

2 dollars

Step-by-step explanation:

5% interest is 1.05

1.05 times 40 equals 42

42-40 equal 2

So 2 dollars

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