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3 years ago

There are 4 angles shown. Which of these are Acute,Right,Obtuse,and Straight angle?

Mathematics

1 answer:

pickupchik [31]3 years ago

6 0

For your problem the second angle is the correct angle

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Multiply. (−9/12) x 4/5 What is the product?

patriot [66]

Answer:

-3x/5

Step-by-step explanation:

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3 years ago

Write the equation in standard form using integers y=2/3x-3

Charra [1.4K]

Answer:2x -3y=9 .

Step-by-step explanation:

7 0

3 years ago

Prove that. 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

Solve the logarithmic equation for x.

aev [14]

Answer:

x = 2 or x = 4

Step-by-step explanation:

Solve: 2 log x = log 2 + log(3x - 4)

log (x^2) = log 2 + log (3x - 4)

log (x^2) = log [2*(3x - 4)]

x^2 = 2*(3x - 4)

x^2 = 6x - 8

x^2 - 6x + 8 = 0

factor:

(x - 2 )(x - 4) = 0

x = 2 or x = 4

4 0

3 years ago

Garth is giving 10 of the action figures from his collection to his brother. If a represents the number of action figures Garth

lawyer [7]

The answer is “a-10”
have a great dayyy!!! :)

4 0

3 years ago