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Darya [45]
3 years ago
7

What is the answer?! Please help!

Mathematics
1 answer:
Step2247 [10]3 years ago
8 0
The notation "sqrt" is shorthand for "square root"

sqrt(-x+10)-x = 2
sqrt(-x+10) = x+2 ... add x to both sides
-x+10 = (x+2)^2 ... square both sides
-x+10 = x^2+4x+4 ... use FOIL
x^2+5x-6 = 0 ... get everything to one side; combine like terms
(x-1)(x+6) = 0 ... factor
x-1=0 or x+6 = 0 ... zero product property
x = 1 or x = -6  .... solve each sub-equation for x

The possible solutions are x = 1 or x = -6. We need to check each of those solutions

Let's check x = 1
sqrt(-x+10)-x = 2
sqrt(-1+10)-1 = 2
sqrt(9)-1 = 2
3-1 = 2
2 = 2
So x = 1 has been confirmed

Let's check x = -6
sqrt(-x+10)-x = 2
sqrt(-(-6)+10)-(-6) = 2
sqrt(6+10)+6 = 2
sqrt(16)+6 = 2
4+6 = 2
10 = 2
The last equation is false, so x = -6 is extraneous. It's not a true solution

----------------------------------

In summary, the value x = 1 is the only solution

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Answer:

\frac{d(cos^{-1}x )}{dx} = \frac{-1}{\sqrt{1-x^2} }

Step-by-step explanation:

Given the differential (d/dx)(cos−1(x)), to find the equivalent formula we will differentiate the inverse function using chain rule as shown below;

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y = cos^{-1} x \\\\taking \ cos\ of\ both\ sides\\\\cosy = cos(cos^{-1} x)\\\\cosy = x\\\\x = cosy\\\\\frac{dx}{dy} = -siny\\

\frac{dy}{dx} = \frac{-1}{sin y}  \\\\from\ trigonometry\ identity,\ sin^{2} x+cos^{2}x = 1\\sinx = \sqrt{1-cos^{2} x}

Therefore;

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Since x = cos y from the above substitute;

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

Hence, \frac{d(cos^{-1}x )}{dx} = \frac{-1}{\sqrt{1-x^2} } gives the required proof

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SUPER EASY PLEASE HELP! 15 POINTS.
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Answer:

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Step-by-step explanation:

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Read 2 more answers
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