Angle RST is a right angle. Angle RSU has a measure of 25°. What is the measure of angle TSU?.

I WILL MARK BRAINEST!!! Someone please help me please!!! I rly need help with question 4, my friend and I both don’t understand

Brums [2.3K]

Answer:

The original number is (i)

The whole amount is (ii)

Step-by-step explanation:

So when finding a number in a percent you wanna see how much that percent will go into that number. So like if I said how many times does 60% go into 200 you would have to see how many times the number 200 would go into 60%. does this make sense? If so this is how you can find your answer.

Sorry if I'm wrong I hope you get this right!! Stay beautiful ❤️

5 0

3 years ago

Please help me with this

Serjik [45]

Answer: you should study, plus give me brainliest please. lol

Step-by-step explanation:

and the other answer might be a bot sending links to websites with malware, please dont .open the link unless you want malware on your device, if its fake. and you cant make websites/ links just to post answers. this is a warning.

5 0

3 years ago

How did the League of Nations' failure to end the Abyssinian Crisis help lead to World War II? The failure showed that even a po

german

Answer:

C - The failure was a starting point for a buildup of aggression around the world.

Step-by-step explanation:

EDGE2021 :-)

Have a nice day! ;)

4 0

3 years ago

Read 2 more answers

What is the square root of 112a^2?

marshall27 [118]

That’s ur answer, your welcome

4 0

3 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