All categories

3 years ago

Help pleaseeeee 20 points

Mathematics

1 answer:

liberstina [14]3 years ago

3 0

It’s the second one!

You might be interested in

10 Write the fraction two different ways.

Cerrena [4.2K]

Answer:

-0.8, -77%

Step-by-step explanation:

These have been rounded up by the way,

I converted it to decimal and percent form

5 0

2 years ago

A = πab/4 solve for b

Studentka2010 [4]

Answer:

4A/ (πa) = b

Step-by-step explanation:

A = πab/4

Multiply each side by 4

4A = 4πab/4

4A = πab

Divide each side by πa

4A/(πa) = πab/ πa

4A/ (πa) = b

8 0

3 years ago

I need help answering all the questions plz

Natali [406]

The width is basically the diameter, and all you have to do is divide by 2 to get the radius.B is the name for base. I think the formula for a triangle is 1/2 bh. and for the circle c=pieD i hope that helped!

4 0

3 years ago

A factory robot recently examined some bulbs, of which 4 were flawed and 26 were not. Considering this data, how many flawed bul

ASHA 777 [7]

Answer:

Step-by-step explanation:

Given that:

Flawed bulbs = 4

Not flawed = 26

Total examined = (flawed + not flawed)

Total examined = (4 + 26) = 30

P(flawed) = flawed / total

P(flawed) = 4 / 30 = 0.1333333

If 15 is examined :

Expected number of flawed bulbs :

(4/30) * 15 = 2

4 0

3 years ago

Dy/dx if y = Ln (2x3 + 3x).

NeTakaya

Answer:

$\frac{6x^2+3}{2x^3+3x}$

Step-by-step explanation:

You need to apply the chain rule here.

There are few other requirements:

You will need to know how to differentiate $\ln(u)$ .

You will need to know how to differentiate polynomials as well.

So here are some rules we will be applying:

Assume $u=u(x) \text{ and } v=v(x)$

$\frac{d}{dx}\ln(u)=\frac{1}{u} \cdot \frac{du}{dx}$

$\text{ power rule } \frac{d}{dx}x^n=nx^{n-1}$

$\text{ constant multiply rule } \frac{d}{dx}c\cdot u=c \cdot \frac{du}{dx}$

$\text{ sum/difference rule } \frac{d}{dx}(u \pm v)=\frac{du}{dx} \pm \frac{dv}{dx}$

Those appear to be really all we need.

Let's do it:

$\frac{d}{dx}\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot \frac{d}{dx}(2x^3+3x)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (\frac{d}{dx}(2x^3)+\frac{d}{dx}(3x))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot \frac{dx^3}{dx}+3 \cdot \frac{dx}{dx})$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot 3x^2+3(1))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (6x^2+3)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{6x^2+3}{2x^3+3x}$

I tried to be very clear of how I used the rules I mentioned but all you have to do for derivative of natural log is derivative of inside over the inside.

Your answer is $\frac{dy}{dx}=\frac{(2x^3+3x)'}{2x^3+3x}=\frac{6x^2+3}{2x^3+3x}$ .

3 0

3 years ago