Which of the following represent the length of a diagonal of this trapezoid?

Mathematics

1 answer:

wariber [46]3 years ago

8 0

Answer:

$\sqrt{\left(a+b\right)^{2}+c^{2}}$

Step-by-step explanation:

The distance between the line joining points A (-b, c) and C (a, o) will be the length of the diagonal of given trapezoid.

Now, $x_{1} = - b,\ y_{1} = c\ and\ x_{2} = a, y_{2} = 0$

Now, using distance formula,

AC = $\sqrt{\left (x_{2}-x_{1}\right )^{2}+\left (y_{2}-y_{1}\right )^{2}}$

AC = $\sqrt{\left(a-(-b) \right)^{2}+\left(0-c\right)^{2}}$

AC = $\sqrt{\left(a+b\right)^{2}+\left(-c\right)^{2}}$

AC = $\sqrt{\left(a+b\right)^{2}+c^{2}}$

Therefore, the length of diagonal AC of trapezoid = $\sqrt{\left(a+b\right)^{2}+c^{2}}$

So, option (4) is the right option.

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Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the area of the region bounded by y = x³ and y = x is approximately 0.267 square units.