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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Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The f

quester [9]

Answer:

It is continuous since $\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$

Step-by-step explanation:

We are given that the function is defined as follows $f(x) = 9-x, x\leq 0$ and $f(x) = 9+12x, x>0$ and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity) x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all $\mathbb{R}$ . So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

$\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$ (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

$\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9$ . This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

$\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9$ . As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.