If the temperature was 25 degrees Celsius and it dropped 32 degrees what temperature is it

Mathematics

2 answers:

n200080 [17]3 years ago

5 0

Answer:

-7 degrees celcius

Step-by-step explanation:

To do this we just need to subtract 32 from 25, indicated by "dropped."

25-32=-7

Hunter-Best [27]3 years ago

3 0

Answer:

-7 degrees Celsius

Step-by-step explanation:

You subtract 25 by 32 and get -7 degrees.

You might be interested in

Need help fast don't know what to do

diamong [38]

Answer:

maybe the answer is

<h2>HOPE IT HELPED </h2>

4 0

3 years ago

Find the expansions of sin⁡5θ and cos⁡5θ in terms of sin⁡θ and cos⁡θ.

qwelly [4]

$\bf \textit{quad-angle identities}\\\\\\

cos(4\theta)=8cos^4(\theta)-8cos^2(\theta)+1
\\\\\\
sin(4\theta)=
\begin{cases}
8sin(\theta)cos^3(\theta)-4sin(\theta)cos(\theta)\\\\
4sin(\theta)cos(\theta)-8sin^3(\theta)cos(\theta)
\end{cases}\\\\
-----------------------------\\\\
$

$\bf sin(5\theta)\iff sin(4\theta + \theta)
\\\\\\
sin(4\theta)cos(\theta)+cos(4\theta)sin(\theta)
\\\\\\\
[4sin(\theta)cos(\theta)-8sin^3(\theta)cos(\theta)]cos(\theta)\\\\ + [8cos^4(\theta)-8cos^2(\theta)+1]sin(\theta)$

$\bf -----------------------------\\\\
cos(5\theta)\iff cos(4\theta + \theta)
\\\\\\
cos(4\theta)cos(\theta)-sin(4\theta)sin(\theta)\\\\\\\
[8cos^4(\theta)-8cos^2(\theta)+1]cos(\theta)\\\\
-
[4sin(\theta)cos(\theta)-8sin^3(\theta)cos(\theta)]sin(\theta)$

distribute... and simplify :)

4 0

3 years ago

HELP IN NUMBER 12 ...!

olasank [31]

The answer is 69.45
hope this helps :)

6 0

3 years ago

Read 2 more answers

Find the measure of arc JK.

allsm [11]

Answer:

lol