Solving [tex]h=vt-2t^2[/tex] for v gives [tex]v=\frac{h}{t}+2t[/tex]
Solving [tex]A=\frac{1}{2}(b_1+b_2)}[/tex] for b1 gives [tex]b_1=2A-b_2[/tex]
Step-by-step explanation:
Given
[tex]h=vt-2t^2[/tex]
To solve for v, we have to isolate v on one side of the equation
[tex]h=vt-2t^2\\Adding\ 2t^2\ on\ both\ sides\\h+2t^2=vt-2t^2+2t^2\\h+2t^2=vt\\Dividing\ both\ sides\ by\ t\\\frac{h+2t^2}{t}=\frac{vt}{t}\\v=\frac{h+2t^2}{t}\\v=\frac{h}{t}+\frac{2t^2}{t}\\v=\frac{h}{t}+2t[/tex]
Given
[tex]A=\frac{1}{2}(b_1+b_2)}[/tex]
To isolate b1 we have to isolate b1
So,
[tex]A=\frac{1}{2}(b_1+b_2)}\\Multiplying\ both\ sides\ by\ 2\\2A=2*\frac{1}{2}(b_1+b_2)}\\2A=b_1+b_2\\Subtracting\ b_2\ from\ both\ sides\\2A-b_2=b_1+b_2-b_2\\b_1=2A-b_2[/tex]
Hence,
Solving [tex]h=vt-2t^2[/tex] for v gives [tex]v=\frac{h}{t}+2t[/tex]
Solving [tex]A=\frac{1}{2}(b_1+b_2)}[/tex] for b1 gives [tex]b_1=2A-b_2[/tex]
Keywords: Variable, Linear equations
Learn more about equations at:
- brainly.com/question/10760452
- brainly.com/question/10882895
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