Find the product of 51 and -2.5.

Is this negative positive or no correlation

Reika [66]

Step-by-step explanation:

No correlation

8 0

2 years ago

I really need help... factor by the simplest form 46xy^5

Ierofanga [76]

The simplest form is 46xy
Since 46 can not be factor out, there is only one x which stay there, and is 5 so the two pairs of y will be canceled and one y left.

5 0

2 years ago

Doing odd jobs in America <br>2000$ How much will it take to earn?<br>I want to know time

kari74 [83]

Answer:

Below:

Step-by-step explanation:

20 Hour Per Week.....

Hope it helps....

It’s Muska...≧◔◡◔≦

3 0

2 years ago

Write a description of how a diagram can help you solve 2×40

nasty-shy [4]

Let me explain like this .Suppose There are 40 people gathered somewhere for attending an event.Each Person has to be given 2 sweets for their arrival or for showing their presence in the event. Now 40 other woman joined them..Instead of getting two sweets each of them got one sweet.

Solution: 2 sweets × 40 people = 1 sweet × 80 people = 80

2nd Way:

There are 40 rooms in a building.In Each room we have to fix two bulbs. So How many bulbs are there in the building.

Solution →→→2 bulbs × 40 room = 80 bulbs

Diagram is depicted below for two parts.

4 0

3 years ago

Read 2 more answers

How do I differentiate this?

disa [49]

Answer:

$\frac{8}{\left(-x+2\right)^2}$ & $x = 1, x = 3$

Step-by-step explanation:

I'm sure you are familiar with the product rule,

If y = u*v => dy/dx = u * dv/dx + v * dy/dx <----- product rule

In this case:

$u=3x+2,\:v=\left(2-x\right)^{-1},\\=>\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

Now remember the sum rule:

$\frac{d}{dx}\left(3x+2\right) = \frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right),\\\frac{d}{dx}\left(3x\right) = 3,\\\frac{d}{dx}\left(2\right) = 0\\\frac{d}{dx}\left(3x+2\right) = 3$

For this second bit we apply the chain rule:

$\frac{d}{dx}\left(\left(2-x\right)^{-1}\right) = -\frac{1}{\left(2-x\right)^2}\frac{d}{dx}\left(2-x\right),\\\frac{d}{dx}\left(2-x\right) = -1,\\\\=> -\frac{1}{\left(2-x\right)^2}\left(-1\right)\\=> \frac{1}{\left(2-x\right)^2}$

If we substitute these values back into the expression... $\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

...we get the following:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)$

The rest is just pure simplification:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}$

Now let's equate this to equal 8 for the second bit and solve for x:

$\frac{8}{\left(-x+2\right)^2}=8,\\\frac{8}{\left(-x+2\right)^2}\left(-x+2\right)^2=8\left(-x+2\right)^2,\\8=8\left(-x+2\right)^2,\\\left(-x+2\right)^2=1,\\x = 1, x = 3$

4 0

2 years ago