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Aleks04 [339]
1 year ago
8

Isolate I for the literal equation V = IR

Mathematics
1 answer:
Anit [1.1K]1 year ago
3 0

In order to isolate any term from an equation we need to clear out any different variable or numbers different from the one we want

V=IR

in this case I is being multiplied by R, so we need to do the opposite which is divide in order to eliminate

V=\frac{IR}{R}

But since this is an equality we must do it on both sides so de equation does not get altered

\frac{V}{R}=\frac{IR}{R}

Then we simplify the equation

I=\frac{V}{R}

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The sides of a right triangle have a relationship as depicted below, where x is the length of the longer leg in inches.
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The equation to represent the area of the triangle would be:
y = 1/2(x²) - (7/2)x

The equation to represent the perimeter of the triangle would be:
y = 3x - 6

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Explanation
The area of a triangle is found using the formula
A = 1/2bh

For our triangle, b = x and h = x-7, so we have:
A = 1/2(x)(x-7)
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We will replace A with y, so we have:
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The perimeter of a triangle is found by adding together all sides, so we have:
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Combining like terms we get:
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Since both equations have y isolated on one side, it will be easy to use substitution to solve the system:

3x - 6 = 1/2(x²) - (7/2)x

It's easier to work with whole numbers, so we will multiply everything by 2:
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We want all of the variables on one side, so we will subtract 6x:
6x - 12 - 6x = x² - 7x - 6x
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When solving quadratics, we want the equation equal to 0, so we will add 12:
-12+12 = x² - 13x + 12
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This is easy to factor, as there are factors of 12 that sum to -13; -12(-1) = 12 and -12+-1 = -13:
0 = (x-12)(x-1)

Using the zero product property, we know that either x-12=0 or x-1=0; therefore x=12 or x=1.

Putting these back into our equation for perimeter (the simplest one) we have:
y = 3(12)-6 = 36-6 = 30; (12, 30);
y = 3(1) - 6 = 3 - 6 = -3; (1, -3)

We cannot have a negative perimeter, so the only viable solution is (12, 30).
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