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IrinaVladis [17]
3 years ago
8

Help meeeeee plllllsssss 10pts + brainliest!!!!

Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0

Answer:

\frac{3*(8e-4f-1)}{8}

I hope this helps. Sorry for a late response. I used

tiger-algebra.com

PEACE!☘️

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1 - 4 + 6^2 ÷ ( 1 + 2 × 2 )​
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Answer:

Step-by-step explanation:

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3 years ago
Can you help please no links or point stealing or I will report
Harlamova29_29 [7]

Answer:

3. 4+2+x and (4+2)+x

4. 7y-10

Step-by-step explanation:

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3 years ago
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Prove that.<br><br>lim Vx (Vx+ 1 - Vx) = 1/2 X&gt;00 ​
faltersainse [42]

Answer:

The idea is to transform the expression by multiplying (\sqrt{x + 1} - \sqrt{x}) with its conjugate, (\sqrt{x + 1} + \sqrt{x}).

Step-by-step explanation:

For any real number a and b, (a + b)\, (a - b) = a^{2} - b^{2}.

The factor (\sqrt{x + 1} - \sqrt{x}) is irrational. However, when multiplied with its square root conjugate (\sqrt{x + 1} + \sqrt{x}), the product would become rational:

\begin{aligned} & (\sqrt{x + 1} - \sqrt{x}) \, (\sqrt{x + 1} + \sqrt{x}) \\ &= (\sqrt{x + 1})^{2} -(\sqrt{x})^{2} \\ &= (x + 1) - (x) = 1\end{aligned}.

The idea is to multiply \sqrt{x}\, (\sqrt{x + 1} - \sqrt{x}) by \displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} so as to make it easier to take the limit.

Since \displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} = 1, multiplying the expression by this fraction would not change the value of the original expression.

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \lim\limits_{x \to \infty} \left[\sqrt{x} \, (\sqrt{x + 1} - \sqrt{x})\cdot \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}\right] \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}\, ((x + 1) - x)}{\sqrt{x + 1} + \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}}\end{aligned}.

The order of x in both the numerator and the denominator are now both (1/2). Hence, dividing both the numerator and the denominator by x^{(1/2)} (same as \sqrt{x}) would ensure that all but the constant terms would approach 0 under this limit:

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \cdots\\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x} / \sqrt{x}}{(\sqrt{x + 1} / \sqrt{x}) + (\sqrt{x} / \sqrt{x})} \\ &= \lim\limits_{x \to \infty}\frac{1}{\sqrt{(x / x) + (1 / x)} + 1} \\ &= \lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + (1/x)} + 1}\end{aligned}.

By continuity:

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \cdots\\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}} \\ &= \cdots \\ &= \lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + (1/x)} + 1} \\ &= \frac{1}{\sqrt{1 + \lim\limits_{x \to \infty}(1/x)} + 1} \\ &= \frac{1}{1 + 1} \\ &= \frac{1}{2}\end{aligned}.

8 0
3 years ago
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1.) Which of the following is closest to 1?
Vaselesa [24]

Answer:

Step-by-step explanation:

98% it is the only percent and it is close to one

4 0
3 years ago
The school that Jaidee goes to is selling tickets to the annual talent show. On the first day of ticket sales the school sold 10
Olenka [21]

The price of a senior citizen ticket  is $8 and the price of a student ticket is$ 10

Step-by-step explanation:

Let the cost of one  seniors citizen ticket be x and  student  be y

First day:

10 seniors citizen tickets and 11 student for a total of $190.

10x + 11y = 190-----------------------------(1)

Second day:

$160 on the second day by selling 5 senior citizen tickets and 12 student tickets.

5x + 12y = 160---------------------------------(2)

Multiply eq(2) by 2

10x + 24y = 320----------------------------(3)

Subtract eq(1) from (3)

10x + 24y = 320

10x + 11y = 190

-----------------------------

0x + 13 y =  130

---------------------------

13y = 130

y = \frac{130}{13}

y = 10-------------------------------(4)

Substituting (4) in (2)

5x +  12(10) =  160

5x + 120 = 160

5x = 160-120

5x = 40

x = \frac{40}{5}

x = 8

5 0
4 years ago
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