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Nitella [24]
4 years ago
6

Simplify sec(arcsec 1/2). A. undefined B. -1/2 C. 1/2 D. 2

Mathematics
2 answers:
kirill [66]4 years ago
8 0
\bf \textit{what is the sine of }90^o?\qquad 1\textit{ then what is the }sin^{-1}(1)?\quad 90^o
\\\\\\
\textit{now then, what is }sin(~~sin^{-1}(1)~~)?\textit{ is just }1\\\\
-------------------------------\\\\

\bf likewise\qquad 
\begin{cases}
cos(~~cos^{-1}(whatever)~~)\implies  whatever\\
sin(~~sin^{-1}(whatever)~~)\implies  whatever\\
tan(~~tan^{-1}(whatever)~~)\implies  whatever\\
cot(~~cot^{-1}(whatever)~~)\implies whatever\\
sec(~~sec^{-1}(whatever)~~)\implies  whatever\\
csc(~~csc^{-1}(whatever)~~)\implies  whatever
\end{cases}

so, surely you know what that is then.
sukhopar [10]4 years ago
3 0
C.1/2 is answer number
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A candy store makes a 13–pound mixture of gummy bears, jelly beans, and gobstoppers. The cost of gummy bears is $1.00 per pound,
bulgar [2K]
X = gummy bears, y = jelly beans, z = gobstoppers
x + y + z = 13
x + 2y + 2z = 20
x = 3y

3y + y + z = 13
4y + z = 13

3y + 2y + 2z = 20
5y + 2z = 20

4y + z = 13....multiply by -2
5y + 2z = 20
-----------------
-8y - 2z = -26
5y + 2z = 20
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-3y = -6
y = 2........2 lbs jelly beans

x = 3y
x = 3(2)
x = 6....6 lbs gummy bears

x + y + z = 13
6 + 2 + z = 13
8 + z = 13
z = 13 - 8
z = 5........5 lbs gobstoppers

so we have : 6 lbs gummy bears (x), 2 lbs jelly beans (y), and 5 lbs gobstoppers (z)




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