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Alina [70]
3 years ago
10

1. Landon is building two separate sandboxes. The length of sandbox A is 5 feet longer than its width. The width of sandbox B is

2 feet less than the width of sandbox A. The length of sandbox B is 2 times its width. If the both sandboxes have the same perimeter, then find the dimensions of each sandbox.
Mathematics
1 answer:
sergejj [24]3 years ago
7 0

Answer:

bX2 plus 5=y

Step-by-step explanation:

thats the

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Write the equation of a line that is parallel to {x=8}x=8x, equals, 8 and that passes through the point {(-3,-2)}(−3,−2)left par
Alexxandr [17]

Answer:

  x = -3

Step-by-step explanation:

The line parallel to x=8 is another vertical line:

  x = constant

To make it go through a point with an x-coordinate of -3, the constant must be -3.

Your line is ...

  x = -3

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4 years ago
What is five kilometers to the nearest tenth of a mile?
MissTica
Are you asking the equation or the answer
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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)
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\textit{now, the "y" in the inverse, is really just g(x)}
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\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
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\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}
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1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
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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
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Answer:

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Step-by-step explanation:

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