(4.27x108) (9.2x10-5)

Math WILL GIVE BRAINLIEST A. How many different outcomes are represented in your tree diagram?

melomori [17]

Step-by-step explanation:

hello just came here to get likes

7 0

2 years ago

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

laiz [17]

In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

8 0

3 years ago

Find the distance between the two points! <br> PLEASE DO THIS ASAP!!<br> no links

Tems11 [23]

Answer:

5

Step-by-step explanation:

The distance is given by sqrt((-3-0)^2+(-3-1)^2)=sqrt(9+16)=5

5 0

3 years ago

Read 2 more answers

At a production process, the produced items are tested for defects. A defective unit is classified as such with probability 0.9,

Snezhnost [94]

Answer:

We use Baye's theorem: P(A)P(B|A) = P(B)P(A|B)

with (A) being defective and

(B) marked as defective

we have to find P(B) = P(A).P(B|A) + P(¬A)P(B|¬A). .......eq(2)

Since P(A) = 0.1 and P(B|A)=0.9,

P(¬A) = 1 - P(A) = 1 - 0.1 = 0.9

and

P(B|A¬) = 1 - P(¬B|¬A) = 1 - 0.85 = 0.15

put these values in eq(2)

P(B) = (0.1 × 0.9) + (0.9 × 0.15)

= 0.225 put this in eq(1) and solve for P(B)

P(B) = 0.4

6 0

3 years ago

Suppose that A, B, and C are invertible matrices of the same

STALIN [3.7K]

Answer:

Step-by-step explanation:

Given that t A, B, and C are invertible matrices of the same

size.

To pr that $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}.$

to pr that $( ABC) (C^{-1}B^{-1}A^{-1})=e$

LS= $( AB(C (C^{-1})B^{-1}A^{-1})\\=ABIB^{-1}A^{-1})\\=A(BB^{-1})A^{-1})\\=AA^{-1})\\=I$

Thus proved

8 0

3 years ago