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3 years ago

What is 9 decimeters equal to _ or_ meters

Mathematics

1 answer:

Sav [38]3 years ago

3 0

9 decimeters. deci = 0.1

9 *0.1 meters.

0.9 meters

9 decimeters = 0.9 meters.

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Which relation is a function?

docker41 [41]

Answer:

The top right grid

Step-by-step explanation:

The other grids have 2 y outputs for one or more x inputs.

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4 years ago

Each of the following statements describes a quadrilateral. Which of the quadrilaterals are NOT parallelograms? (Pick 3)

Tanya [424]

Answer:

Quadrilaterals 1, 2 and 4 does not represents a parallelogram.

Step-by-step explanation:

We know that, a parallelogram is a simple quadrilateral that have two opposite sides congruent.

Now, according to the options:

1. A parallelogram does not have congruent diagonals without having right angles.

2. A parallelogram does not have consecutive sides equal without being a rhombus.

3. A rectangle is a parallelogram which have both the diagonals of equal length. So, this quadrilateral might represents a rectangle, which is a parallelogram.

4. No two opposite angles of a parallelogram are right angles without the quadrilateral being a rectangle.

5. A rhombus is a parallelogram which have both the diagonals bisecting each other perpendicularly. So, this quadrilateral might represents a rhombus, which is a parallelogram.

Hence, we see that options 1, 2 and 4 does not represents a parallelogram.

5 0

3 years ago

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Determine a series of transformations that would map Figure P onto Figure Q.

pav-90 [236]

Answer: Rotate 280 degrees

Step-by-step explanation:

Each 90 degree rotation makes the figure go into the next quadrant of the graph. So, going clockwise, you have to move the figure three quadrants to get Figure P to the position of Figure Q. Thus, 3 x 90 = 280 degrees. There is only one transformation required.

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3 years ago

Please help with this question

Mars2501 [29]

To find out one book you have to divide 65 by 5

65/5 = 13

The answer is $13 per book

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3 years ago

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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

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