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saveliy_v [14]
3 years ago
14

3(-5)(-1)(-2) evaluate please thank you :D

Mathematics
2 answers:
user100 [1]3 years ago
8 0

Answer:

-30

Step-by-step explanation:

makvit [3.9K]3 years ago
4 0

Answer:

the answer is -30

Step-by-step explanation:

this just means to multiple from left to right

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I need help with this Adding Fractions problem. If you answer and fill out this whole page I will love you for the rest of my li
zhannawk [14.2K]

Answer:

omigosh hope she gets better i will look at the pdf and solve

Step-by-step explanation:

okay so for most of them you can just multiply by the denominator to turn it into a regular number (to get it out of fraction form) and then use basic algebra to solve for x. I 'll write the answers to problems 1-9 in the comments

8 0
3 years ago
The FVHS football team is trying to raise money to buy new equipment so the players are selling “Bengal Cards” for $20 each. If
Sav [38]

Answer:

BANKS LOL????????

Step-by-step explanation:

really? oh wait....... I looked it up

4 0
3 years ago
*PLEASE HELP* *ANY IRRELEVANT ANSWERS WILL BE REPORTED AS SPAM*
drek231 [11]

Answer:

67.307

Step-by-step explanation:

30×25=750

100×80=8000

750+8000=8750

8750/130=67.307... total hits divided by total games

8 0
2 years ago
Read 2 more answers
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
3 years ago
Mateo wants to make a toy sailboat. He has two pieces of wood to choose from.
jeka94

Answer:

wut is it

Step-by-step explanation:

8 0
3 years ago
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