anonymous
  • anonymous
The area of a parallelogram is A = bh. Solve for h (height) in this formula.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
You have h defined in the equation, now you must isolate it.
Jamierox4ev3r
  • Jamierox4ev3r
well if a=bh, then you divide both sides by b, which gives you \(\large\frac{a}{b}=h\), or \(h=\large\frac{a}{b}\). Does that make sense? @mathematicshead ? Next problem I want you to try on your own. :)
anonymous
  • anonymous
& hows that @Anickyan

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Jamierox4ev3r
  • Jamierox4ev3r
well @Anickyan has stated, h is right there in the equation. it is not 2h, h^2, just h. all you had to do was get it by itself
anonymous
  • anonymous
What Jamierox4ev3r said. Try to do it with this equation, just to practice a bit: \[2a + 3 = \frac{ h }{ b } - 5\] Solve for h
Jamierox4ev3r
  • Jamierox4ev3r
hello @mathematicshead ? what should you do to get the fraction \(\Large\frac{h}{b}\) by itself? Think now, and whatever you do must be done to both sides.
anonymous
  • anonymous
i dont understand any of this at all, thats why im on here
anonymous
  • anonymous
You should have listened in class. And if your teacher is assigning you this work, and haven't taught you it yet, you'd probably better talk to him/her.
Jamierox4ev3r
  • Jamierox4ev3r
well think about how you would solve a regular equation. When faced with an equation, just see if you can add or subtract the sides :) for example : x+2=8 -2 -2 ------- x=6
Jamierox4ev3r
  • Jamierox4ev3r
@mathematicshead watch my process carefully and see if you understand: \(2a+3=\large\frac{h}{b}-5\) +5 +5 ----------------- 2a+8=\(\frac{h}{b}\) multiply \(\large\frac{h}{b}\) by b and everything by b and you get : 2ab+8b=h
Jamierox4ev3r
  • Jamierox4ev3r
hey you guys there? O_O @mathematicshead and @Anickyan ?
anonymous
  • anonymous
.................
anonymous
  • anonymous
Yes. mathematicshead, this is really basic algebra. How old are you, and what grade are you in?
anonymous
  • anonymous
@mathematicshead ^

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