Answer:
12 cm²
Step-by-step explanation:
The area (A) of a trapezoid is calculated using the formula
A = h(a + b)
where h is the height and a, b the parallel bases
here h = 3, a = 3 and b = 5, hence
A = × 3 (3 + 5) = 1.5 × 8 = 12 cm²
See attachment for math work and answer.
<h3>Answer:</h3>
<h3>Explanation:</h3>
It can work well to consider the function in parts. Define the following:
... a(x) = (1/2)ln(x^2+3)
... b(x) = x(4x^2-1)^3
Then the derivatives of these are ...
... a'(x) = (1/2)·1/(x^2 +3)·2x = x/(x^2+3)
... b'(x) = (4x^2 -1)^3 + 3x(4x^2 -1)^2·8x = (4x^2 -1)^2·(4x^2 -1 +24x^2)
... = (4x^2 -1)^2·(28x^2 -1)
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<em>Putting the parts together</em>
f(x) = a(x)/b(x)
f'(x) = (b(x)a'(x) -a(x)b'(x))/b(x)^2 . . . . . rule for quotient of functions
Substituting values, we have
... f'(x) = (x(4x^2 -1)^3·x/(x^2 +3) -(1/2)ln(x^2 +3)·(4x^2 -1)^2·(28x^2 -1)) / (x(4x^2 -1)^3)^2
We can cancel (4x^2 -1)^2 from numerator and denominator. We can also eliminate fractions (1/2, 1/(x^2+3)). Then we have ...
... f'(x) = 2x^2(4x^2 -1) -(x^2 +3)ln(x^2 +3)·(28x^2 -1)/(2x^2·(x^2 +3)(4x^2 -1)^4))
Simplifying a bit, this becomes ...
... f'(x) = (8x^4 -2x^2 -ln(x^2 +3)·(28x^4 +83x^2 -3))/(2x^2·(x^2 +3)(4x^2 -1)^4))
Let's solve for c. Step by step.<span><span>a<span>(<span>b−c</span>)</span></span>=d</span>Step 1: Add -ab to both sides.<span><span><span><span>ab</span>−<span>ac</span></span>+<span>−<span>ab</span></span></span>=<span>d+<span>−<span>ab</span></span></span></span><span><span>−<span>ac</span></span>=<span><span>−<span>ab</span></span>+d</span></span>Step 2: Divide both sides by -a.<span><span><span>−<span>ac</span></span><span>−a</span></span>=<span><span><span>−<span>ab</span></span>+d</span><span>−a</span></span></span><span>c=<span><span><span>ab</span>−d</span>a</span></span>Answer:<span>c=<span><span><span>ab</span>−d/</span><span>a
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<h3>
Answer: 1.3 seconds</h3>
Explanation:
Replace h with 0. Solve for t.
It takes about 1.3 seconds for the balloon to hit the ground.