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Gnom [1K]
3 years ago
6

Solve the area formula for h

Mathematics
1 answer:
White raven [17]3 years ago
4 0

The answer is C: h = \frac{2A}{b}

Starting with the given formula for the area, A = 1/2hb, the first step is to isolate h by using the multiplicative inverse of b, which is \frac{1}{b} on both sides of the equation:

A(\frac{1}{b})  =  \frac{1}{2}hb* (\frac{1}{b})

The result will be:

\frac{A}{b} = \frac{1}{2}h

The last step is to use the multiplicative inverse of \frac{1}{2}, which is 2 or \frac{2}{1} to further isolate the variable <em>h </em>:

\frac{A}{b} (\frac{2}{1}) = (\frac{2}{1}) \frac{1}{2} h

The formula for h will be:

h = \frac{2A}{b}  which is the option C in your question.

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Jessica has a circular birdbath in her backyard with a radius of 1.25 feet. She wants to place a decorated border around the cir
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Answer:

The total cost is <em>28.26.</em>

Step-by-step explanation:

The radius of the birdbath is <em>1.25 feet.</em>

The cost per foot is <em>3.6</em> .

<em>The formula for perimeter of circle is given by,</em>

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Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integ
liq [111]

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

  • Mx(x,y) = d/dx 8x²y = 16xy
  • Ly(x,y) = d/dy 7y²x = 14xy

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2

We conclude that the line integral is 1/2

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3 years ago
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