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3 years ago

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the length of a rectangle is 3 units shorter than one -third of the width, x. Which expression represents the perimeter of the r

ectangle? A) 8/3x-6, B) 2/3x -4, C) 2/3x -8, D) 8/3x -2.

1 answer:

Mariana [72]3 years ago

8 0

Answer:

A) $\frac{8}{3} x - 6$

Step-by-step explanation:

Let the width of the rectangle be x units.

$\therefore \: length \: of \: rectangle = \frac{1}{3} x - 3 \\ Perimeter \: of \: rectangle = 2(l + w) \\ = 2 \bigg( \frac{1}{3} x - 3 + x \bigg)\\ = 2 \bigg( x + \frac{1}{3} x - 3 \bigg)\\ = 2 \bigg( \frac{3x + x}{3} - 3 \bigg)\\ = 2 \bigg( \frac{4x}{3} - 3 \bigg) \\ = 2 \bigg( \frac{4x}{3} - 3 \bigg) \\ = \frac{2 \times 4x}{3} -2 \times 3 \\ \purple{ \bold{{Perimeter \: of \: rectangle = \frac{8}{3} x - 6}}}$

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Solve for x for 1 and 2

taurus [48]

Answer:

1. 37.5

2. 8

Step-by-step explanation:

1. the smaller triangle and the bigger triangle are similar, so the sides are also similar.

this means that 16/46 = 20/(x+20)

then, just solve that for x

16/46 = 20/(x+20)

divide both sides by 4

4/46 = 5/(x+20)

multiply both sides by (x + 20)

(x+20) * 4/46 = 5

multiply both sides by 46

4x + 80 = 230

subtract 80

4x = 150

divide 4

x = 37.5

2. do the same thing ( still similar triangles)

the equation would be (3x + 1)/10 = (7x - 1)/22

solve

22(3x + 1) = 10(7x - 1)

66x + 22 = 70x - 10

32 = 4x

x = 8

3 0

3 years ago

Evaluate the square root of x+7 when x equals 9. Write the answer in simplified radical form.

melomori [17]

sqr of 4^2 would be something like 4

6 0

3 years ago

What is 8 21/40 I don't understand how to do it

yaroslaw [1]

Hey, there !

If you want to know it as a "improper fraction" , you can

Multiply the denominator from the front number
Whatever the denominator and front number is , you add it to the numerator

So, for this problem we first have the multiply the front number from the denominator

Like: 8 (40)
That gives us the answer of: 320

Now we have to add 320 to 21

Like: 320 + 21

That gives us the answer of:341

We keep our denominator (which is "40" )

Your improper fraction would be:

341 / 40

Good luck on your assignment and enjoy your day!

~MeIsKaitlyn:)

3 0

3 years ago

Read 2 more answers

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).

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3 years ago

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natali 33 [55]

Answer:

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