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3 years ago

A square with an area of 81 cm^2 is rotated to form a cylinder. What is the height of the cylinder?

1 answer:

7 0

The area of a square is just:

A=s^2 so

s^2=81

s=9 so if this square is rotated about its center you will get a cylinder with a radius of 4.5 cm and a height of 9 cm.

So the height is 9 cm

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A pair of jeans that normally sells for $35 is in sale for 20%. Find the sale price of the jeans. Then find the total cost of th

never [62]

35×.2= 7 dollars is being taken away from the sale.

35-7= 28 dollars which is the new cost.

28×.06=1.68
28+1.68=29.68 final cost of jeans

or you can do 28×1.06= 29.68 final cost of jeans.

7 0

3 years ago

HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider<im

GrogVix [38]

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for $f:\mathbb R^2\to\mathbb R$ , we have the partial derivative with respect to $x$ defined as

$\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h$

The derivative at (0, 0) is then

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h$

• By definition of $f$ , $f(0,0)=0$ , so

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}$

• Expanding the tangent in terms of sine and cosine gives

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}$

• Introduce a factor of $g(h,0)$ in the numerator, then distribute the limit over the resulting product as

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h$

• The first limit is 1; recall that for $a\neq0$ , we have

$\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1$

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h$

and this is exactly the partial derivative of $g$ with respect to $x$ .

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)$

For the same reasons shown above,

$\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)$

(b) To show that $f$ is differentiable at (0, 0), we first need to show that $f$ is continuous.

• By definition of continuity, we need to show that

$\left|f(x,y)-f(0,0)\right|$

is very small, and that as we move the point $(x,y)$ closer to the origin, $f(x,y)$ converges to $f(0,0)$ .

We have

$\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}$

The first expression in the product is bounded above by 1, since $|\sin(x)|\le|x|$ for all $x$ . Then as $(x,y)$ approaches the origin,

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0$

So, $f$ is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which $f(x,y)$ changes as we move the point $(x,y)$ closer to the origin, given by

$\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,$

approaches 0.

Just like before,

$\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}$

and this converges to $g(0,0)=0$ , since differentiability of $g$ means

$\displaystyle \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)}{\sqrt{x^2+y^2}}=0$

So, $f$ is differentiable at (0, 0).

3 0

3 years ago

Can u help solve this

Klio2033 [76]

Answer:

3 or 6/2

Step-by-step explanation:

4--2 (you add there is a subtraction of a negative) over or divided by 3-1

4 0

3 years ago

20 POINTS!!! PLZ ANSWER!! On the moon, the time, in seconds, it takes for an object to fall a distance, d, in feet, is given by

scoundrel [369]

Answer:

Part a) $f(2)=1.57\ sec$

It takes 1.57 seconds for an object to fall a distance of 2 feet

Part b) see the explanation

Part c) $2,165.43\ sec$

Step-by-step explanation:

Let

f(d) -----> the time in seconds it takes for an object to fall

d -----> distance in feet

we have

$f(d)=1.11\sqrt{d}$

Part a): Determine f(2) and explain what it represents

we know that

f(2) represent the time in seconds it takes for an object to fall a distance of 2 feet

For $d=2\ ft$

substitute in the function above and solve for f(2)

$f(2)=1.11\sqrt{2}$

$f(2)=1.57\ sec$

therefore

It takes 1.57 seconds for an object to fall a distance of 2 feet

Part b) The imbrium basin is the largest basin on the moon. A reasonable domain for the height above the lowest point in the basin is given by {d|0 ≤ d ≤ 3805774}

What does this tell you about the basin?

The height of the basin is greater than or equal to 0 ft and less than or equal to 3,805,774 ft

The maximum height of the basin is 3,805,774 ft

Part c) How long would it take a rock to drop from the rim to the bottom of the basin?

we know that

The distance from the rim to the bottom of the basin is equal to the maximum height of the basin

$d=3,805,774\ ft$

substitute the value of d in the function f(d)

$f(d)=1.11\sqrt{3,805,774}$

$f(d)=2,165.43\ sec$

3 0

3 years ago

James bought a new sweatshirt from Foot Locker for $39.98 with tax of 7% that was added at the register. James gave the cashier

Veseljchak [2.6K]

Answer: 2.2214

Step-by-step explanation:

0.07 x 39.98 = 2.7986 (is tax)

2.7986 + 39.98 = 42.7786

45.00 - 42.7786 = 2.2214

6 0

2 years ago