Answer:
A
Step-by-step explanation:
The projectile motion of Jenny's dive can be represented by 
where a is acceleration, v_0 is velocity, t is time, and h_0 is initial height.
Jenny's vertical height above the water is represented by 
. 10 is h_0 and is Jenny's initial height.
Let's solve your equation step-by-step.<span><span><span>2x</span>+<span>3<span>(<span>x−4</span>)</span></span></span>=<span>x−20</span></span>Step 1: Simplify both sides of the equation.<span><span><span>2x</span>+<span>3<span>(<span>x−4</span>)</span></span></span>=<span>x−20</span></span><span>Simplify: (Show steps)</span><span><span><span>5x</span>−12</span>=<span>x−20</span></span>Step 2: Subtract x from both sides.<span><span><span><span>5x</span>−12</span>−x</span>=<span><span>x−20</span>−x</span></span><span><span><span>4x</span>−12</span>=<span>−20</span></span>Step 3: Add 12 to both sides.<span><span><span><span>4x</span>−12</span>+12</span>=<span><span>−20</span>+12</span></span><span><span>4x</span>=<span>−8</span></span>Step 4: Divide both sides by 4.<span><span><span>4x/</span>4</span>=<span><span>−8/</span>4</span></span><span>x=<span>−2</span></span>Answer: x=−2
I think the answer is 25.5. Although, I don't know how you would have half a bike unless there was a 3 wheel bike which would make the answer 24.
The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span> =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).
Answer:
5
Step-by-step explanation:
its 5
