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yarga [219]
3 years ago
15

Pls help quick

Mathematics
1 answer:
Inga [223]3 years ago
7 0
For the second one, $5.00 is the change
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A point moves along the curve y = √ x in such a way that the y-component of the position of the point is increasing at a rate of
Eduardwww [97]

Answer:

The value of x component changes at a rate of \frac{dx}{dt}=4\sqrt{x} units per second

Step-by-step explanation:

We are given that y=x^{\frac{1}{2}}

Differentiating on both sides with respect to time we get

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

It is given that \frac{dy}{dt}=2units/sec

Solving for \frac{dx}{dt} we get

\frac{dx}{dt}=\frac{dy}{dt}\times 2\sqrt{x}\\\\\frac{dx}{dt}=4\sqrt{x}

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A car is moving 70 miles per hour. how far does it travel in 16secs
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6 0
3 years ago
Help please!! How do I solve this
egoroff_w [7]
This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
 2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0  We will replace expression ( 2 on x) with variable t => 2 on x=t  =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 =>     t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 =>                   2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution. 
Check 2:   2 on (2*0) - 5* 2 on 0 + 4=0 => 2 on 0 - 5 * 1 + 4=0 => 
1-5+4=0 => 0=0  Identity provin solution.
5 0
3 years ago
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