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3 years ago

HELP PLEASE!!! ANSWERS WITH LINKS WILL BE DELETED AND REPORTED!!!

Mathematics

2 answers:

Mariana [72]3 years ago

7 0

what he/she said BAHAHAHAHAA

charle [14.2K]3 years ago

3 0

Answer:

1,066cm

Step-by-step explanation:

First, subtract. 55 - 31 = 24

Then, multiply. 24 × 16 = 384

Next, multiply again. 22 × 31 = 682

Lastly, add. 384 + 682 = 1,066

Have a beautiful day!

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A robot can complete 8 tasks in ⅚ hour. Each task takes the same amount of time. How long does it take the robot to complete one

nikitadnepr [17]

Answer:

1 task = 6.25 minutes

The robot can complete 9.6 task in 1 hour

Step-by-step explanation:

Number of task = 8

Time taken = 5/6 of 1 hour

5/8 of 1 hour = 5/6 × 60 minutes

= 300/6

= 50 minutes

8 task takes 50 minutes

1 task = 50 minutes / 8 task

= 6.25 minutes

1 task = 6.25 minutes

x task = 60 minutes

1 : 6.25 = x : 60

1/6.25 = x / 60

1*60 = 6.25*x

60 = 6.25x

x = 60/6.25

= 9.6

x= 9.6

The robot can complete 9.6 task in 1 hour

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).

3 0

3 years ago

Simplify 3x^2+ 13x – 10 9x^2 - 4

svetlana [45]

Answer:

The top one is (3x−2)(x+5)

the bottom one is (3x−2)(3x+2)

Step-by-step explanation:

4 0

3 years ago

Determine the time needed to deliver 72 papers at a rate of 9 papers in 18 minuets

monitta

Hey there,

Question : Determine the time needed to deliver 72 papers at a rate of 9 papers in 18 minuets

Answer :
9 papers = 18 min
1 paper = 18 / 9
= 2 min
72 papers = 2 x 72
= 144 mins

Hope this helps :))

~Top♥

3 0

3 years ago

Points R, S, and T are collinear, and S is between R and T, in addition, RS = 2x-4, ST = 3x+2, and RT = 13. Find the value of RS

Alex17521 [72]

Answer:

RS = 2

Step-by-step explanation:

Points R, S, and T are collinear, and S is between R and T, in addition, RS = 2x-4, ST = 3x+2, and RT = 13. Find the value of RS

Hence:

RS + ST = RT

2x - 4 + 3x + 2 = 13

2x + 3x - 4 + 2 = 13

5x - 2 = 13

5x = 13 + 2

5x = 15

x = 15/5

x = 3

To find the value of RS

RS = 2x-4

RS = 2 × 3 - 4

RS = 6 - 4

RS = 2

5 0

3 years ago