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Sergio [31]
3 years ago
5

Determine whether it is a solution to 2+3x≥8​

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
7 0
2 + 3x =8 *i can’t put the sign*
subtract 2 on both sides
3x =6
Diverse 3 on both sides
x=2

yes it’s a solution
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Answer:

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3 years ago
D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx
sineoko [7]

Answer:

\frac{1}{2\sqrt{5} }

Step-by-step explanation:

Let, \text{sin}^{-1}(\frac{x}{6}) = y

sin(y) = \frac{x}{6}

\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})

\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx} ---------(1)

\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}

\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}

cos(y) = \sqrt{1-\text{sin}^{2}(y) }

          = \sqrt{1-(\frac{x}{6})^2}

          = \sqrt{1-(\frac{x^2}{36})}

Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

Or \frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

                   =\frac{1}{2\sqrt{5}}

4 0
3 years ago
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sergiy2304 [10]

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Step-by-step explanation:

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Vaselesa [24]
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3 0
3 years ago
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An employment agency specializing in temporary construction help pays heavy equipment operators $138 per day and general laborer
prisoha [69]

Answer:

L = 2,

O = 31

Step-by-step explanation:

First using the givens, write out some formulas.

Heavy equipment operators = O

General laborers = L

  • O + L = 33
  • 138O + 87L = 4044

Now use substitution, it doesn't matter which one you substitute.

  • O = -L +33
  • 138( -L + 33 ) + 87L = 4044

Distribute and solve for L.

  • -138L + 4554 + 87L = 4044
  • -138L - 87L = 4044 - 4554
  • -225L = -510
  • L = -510/-225
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L = 2, meaning number of laborers employed is 2.

Now plug 2 = L into one of the original equations.

  • O + L = 33
  • O + 2 = 33
  • O = 33 - 2
  • O = 31

O = 31, meaning number of heavy equip. operators employed is 31.

4 0
3 years ago
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