Solve for d small diameter [tex]A=\frac{\pi *D^{2} }{4} – \frac{\pi *d^{2} }{4}[/tex]

Question

Solve for d small diameter
[tex]A=\frac{\pi *D^{2} }{4} – \frac{\pi *d^{2} }{4}[/tex]

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Trúc Chi 1 year 2021-09-04T14:50:34+00:00 2 Answers 13 views 0

Answers ( )

  1. Answer:

    [tex] \boxed{ \tt{d = \sqrt{D^{2} -\frac{4A}{\pi}} }}[/tex]

    Step-by-step explanation:

    [tex]if \: A=\frac{\pi *D^{2} }{4} – \frac{\pi *d^{2} }{4} \\ then \to \\ A = \frac{\pi}{4} (D^{2} – {d}^{2} ) \\ (D^{2} – {d}^{2} ) \pi = 4A \\ D^{2} – {d}^{2} = \frac{4A}{\pi} \\ {d}^{2} = D^{2} -\frac{4A}{\pi} \\ d = \sqrt{D^{2} -\frac{4A}{\pi}} [/tex]

    0
    2021-09-04T14:51:54+00:00

    Answer:

    [tex]\boxed{\text{\Large \sqrt{-1.27324A+D^2}$}}[/tex]

    Step-by-step explanation:

    [tex]\displaystyle A=\frac{\pi \times D^2}{4} – \frac{\pi \times d^2}{4}[/tex]

    [tex]\displaystyle A=0.785398D^2 – 0.785398d^2[/tex]

    Solve for d

    [tex]\displaystyle A-0.785398D^2= – 0.785398d^2[/tex]

    [tex]\displaystyle \frac{A-0.785398D^2}{- 0.785398} = d^2[/tex]

    [tex]-1.27324A+D^2=d^2[/tex]

    [tex]\displaystyle \sqrt{-1.27324A+D^2} =d[/tex]

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