11th grade geometry:

Mathematics

1 answer:

Aliun [14]3 years ago

3 0

Answer:

The perimeter is 72 units and the area is 149 square units.

Step-by-step explanation:

$\triangle SBA$ has coordinates $S(15,-8),B(-2,21)$ and $A(0,0)$

Using the distance formula.........

Length of side $SB = \sqrt{(15+2)^2+(-8-21)^2}= \sqrt{17^2+(-29)^2}= \sqrt{1130}$

Length of side $BA= \sqrt{(-2)^2+(21)^2}= \sqrt{445}$

Length of side $AS =\sqrt{(15)^2+(-8)^2}=\sqrt{289}=17$

So, the perimeter of the triangle will be: $(SB+BA+AS)= \sqrt{1130}+ \sqrt{445}+17 =71.71... \approx 72$ units. (Rounded to the nearest unit)

The height of the triangle for the corresponding base $SB$ is 8.89 units.

Formula for the Area of triangle, $A= \frac{1}{2}(base\times height)$

So, the area of the $\triangle SBA$ will be: $\frac{1}{2}(\sqrt{1130}\times 8.89)= 149.42... \approx 149$ square units. (Rounded to the nearest unit)

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For the function given below, find a formula for the Riemann sum obtained by dividing the interval (0, 3) into n equal subinterv

Viktor [21]

Splitting up [0, 3] into $n$ equally-spaced subintervals of length $\Delta x=\frac{3-0}n = \frac3n$ gives the partition

$\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right]$

where the right endpoint of the $i$ -th subinterval is given by the sequence

$r_i = \dfrac{3i}n$

for $i\in\{1,2,3,\ldots,n\}$ .

Then the definite integral is given by the infinite Riemann sum

$\displaystyle \int_0^3 2x^2 \, dx = \lim_{n\to\infty} \sum_{i=1}^n 2{r_i}^2 \Delta x \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac6n \sum_{i=1}^n \left(\frac{3i}n\right)^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3} \sum_{i=1}^n i^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3}\cdot\frac{n(n+1)(2n+1)}6 = \boxed{18}$