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3 years ago

1 point Solve: 0.75 (x + 200) = 150.5 (2 – 28) type your answer...

Mathematics

1 answer:

adelina 88 [10]3 years ago

4 0

Answer:

$\huge \: x = - 5417.33$

Step-by-step explanation:

$0.75 (x + 200) = 150.5 (2 – 28) \\$

Expand the terms and simplify

That's

$0.75x + 150 = 301 - 4214 \\ 0.75x + 150 = - 3913$

Subtract 150 from both sides of the equation

$0.75x + 150 - 150 = - 3913 - 150 \\ 0.75x = - 4063$

Divide both sides by 0.75

$\frac{0.75x}{0.75} = - \frac{4063}{0.75} \\ x = - 5417.333333$

We have the final answer as

Hope this helps you

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Iliana cubed the number of flowering plants in her garden, then added 2 vegetable plants. Let p represent the original number of

adelina 88 [10]

When she cubed the number of flowering plants that means p^3 so the answer is not A or B
now take a look at C or D
both are p^3 + 2 which is right now lets solve it:
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so the answer is D :)))
i hope this is helpful
have a nice day

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3 years ago

Read 2 more answers

The three lines given below form a triangle. Find the coordinates of the vertices of the triangle.

LenKa [72]

Answer:

(3, 5), (1, 2) and (5, 1)

Step-by-step explanation:

Make three systems with pairs of lines and solve them to work out the vertices.

1) Line 1 and line 2

-3x + 2y = 1
2x + y = 11

Double the second equation and subtract equations:

-3x + 2y - 2(2x + y) = 1 - 2(11)
-3x - 4x = 1 - 22
-7x = - 21
x = 3

Find y:

2*3 + y = 11
6 + y = 11
y = 11 - 6
y = 5

The point is (3, 5)

2) Line 1 and line 3

-3x + 2y = 1
x + 4y = 9

Triple the second equation and add up equations:

-3x + 2y + 3(x + 4y) = 1 + 3(9)
2y + 12y = 1 + 27
14y = 28
y = 2

Find x:

x + 4*2 = 9
x + 8 = 9
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The point is (1, 2)

3) Line 2 and line 3

2x + y = 11
x + 4y = 9

Double the second equation and subtract the equations:

2x + y - 2(x + 4y) = 11 - 2(9)
y - 8y = 11 - 18
- 7y = - 7
y = 1

Find x:

x + 4*1 = 9
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2 years ago

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2 years ago

How are the two angles related?

wlad13 [49]

Answer:

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Step-by-step explanation:

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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.

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3 years ago