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Kaylis [27]
3 years ago
14

Consider the line y=1 x -1 and the point P=(2,0).

Mathematics
1 answer:
Montano1993 [528]3 years ago
8 0

Answer:

Step-by-step explanation:

d(x)=√((x-2)2+(y-0)²)

=√((x-2)²+y²)

=√((x-2)²+(x-1)²)

=√(x²-4x+4+x²-2x+1)

=√(2x²-6x+5)

D=d²(x)=2x²-6x+5

b.\\\frac{dD}{dx}=4x-6\\\frac{dD}{dx}=0,gives \\4x-6=0\\x=\frac{3}{2}\\\frac{d^2D}{dx^2}=4 >0 at x=\frac{3}{2}\\so~D~or~d^2~or~d~is~minimum~at~x=\frac{3}{2}\\so~y=1(\frac{3}{2} )-1=1/2=0.5\\so~nearest~point~is~(1.5,0.5)

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3 years ago
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Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step.  The equation is

8-3\sqrt[5]{x^3}=-7

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time.  First thing is to subtract 8 from both sides:

-3\sqrt[5]{x^3}=-15

The goal is to isolate the term with the x in it, so that means that the -3 has to go.  Divide it away on both sides:

\sqrt[5]{x^3}=5

Let's rewrite that radical into exponential form:

x^{\frac{3}{5}}=5

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely.  On the right, let's rewrite that back in radical form to solve it easier:

x=\sqrt[3]{5^5}

Let's group that radicad into groups of 3's now to make the simplifying easier:

x=\sqrt[3]{5^3*5^2} because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

x=5\sqrt[3]{5^2} which is the same as:

x=5\sqrt[3]{25}

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Slope intercept form that passes though the point (-2,-3) and has a slope of 5/2
olganol [36]

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