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2 years ago

The area of a rectangle is 2x (length + breadth) true or false

Mathematics

2 answers:

erma4kov [3.2K]2 years ago

3 0

Answer:

false

Step-by-step explanation:

to get the area of a rectangle u need to multiple length and width.

Hoochie [10]2 years ago

3 0

Answer:

false

Step-by-step explanation:

the given expression describes the perimeter.

the area is

length × width (or breadth or depth or whatever word applies best for the second dimension in a certain scenario)

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3 years ago

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Find the maximum rate of change of f at the given point and the direction in which it occurs. f(p, q) = 8qe−p + 4pe−q, (0, 0) ma

Pachacha [2.7K]

Answer:

Step-by-step explanation:

$Given \ f (p,q) = 8qe^{-p} + 4pe^{-q} \ where \ P_o(0,0) \\ \text{thus the gradient of f is: } \\ \\ \bigtriangledown f(p,q) = \Big(\dfrac{\partial f}{\partial p}, \dfrac{\partial f}{\partial q} \Big) \\ \\ \dfrac{\partial f}{\partial p} = -8qe^{-p} + 4pe^{-q} \\ \\ \dfrac{\partial f}{\partial p} = 8qe^{-p} - 4pe^{-q} \\ \\ Then: \bigtriangledown f(p.q) = (-4qe^{-p}+ 8qe^{-q}, 4qe^{-p}- 8qe^{-q}) \\ \\ f(0,0) = (-4*(0)e^{-0}+ 8*(0)e^{-(0)}, 4*(0)e^{-(0)}- 8*(0)e^{-(0)}) \\ \\ = (0+8,4-0) = (8.4)$

$\Big| \Big | \bigtriangledown f(0,0) = \sqrt{(8)^2+4^2} \\ \\ = \sqrt{64+16} \\ \\ = \sqrt{80}$

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3 years ago

What is the volume of the triangular pyramid

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I hope this helps you

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4 years ago

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

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Wher is said drawing?

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