All categories

3 years ago

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!☘️

You might be interested in

1 - 4 + 6^2 ÷ ( 1 + 2 × 2 )

solniwko [45]

Answer:

Step-by-step explanation:

957 6547

4 0

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:

3 0

3 years ago

Read 2 more answers

Prove that.<br><br>lim Vx (Vx+ 1 - Vx) = 1/2 X>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

Read 2 more answers

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