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2 years ago

Mark is painting the ceiling of a square room. If the area he paints is 400 sq. feet, what are the dimensions of the room?

Mathematics

1 answer:

Sedbober [7]2 years ago

7 0

Answer:

20 by 20

Step-by-step explanation:

square root of 400 is 20

20 times 20 = 400

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What is the solution set for this linear-quadratic system of equations?

Lostsunrise [7]

Y - x - 3 = 0
y = x + 3

y = x^2 - x - 12
x + 3 = x^2 - x - 12
x^2 - x - 12 - x - 3 = 0
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0

x - 5 = 0 y = x + 3
x = 5 y = 5 + 3
y = 8

x + 3 = 0 y = x + 3
x = -3 y = -3 + 3
y = 0

so ur solutions are : (5,8) and (-3,0)

3 0

3 years ago

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You're playing a game where you defend your village from an orc invasion. There are 333 characters (elf, hobbit, or human) and 5

antiseptic1488 [7]

Answer:

8/15

Step-by-step explanation:

khan acdamy

7 0

3 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

I would like some help please

wlad13 [49]

Answer:

it would have to be c....

Step-by-step explanation:

4 0

3 years ago

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D= 2 - 6c <br> 1%2c - c = 1

omeli [17]