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Lesechka [4]
3 years ago
13

11th grade geometry:

Mathematics
1 answer:
Aliun [14]3 years ago
3 0

Answer:

<em>The perimeter is 72 units and the area is 149 square units.</em>

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.   <em>(Rounded to the nearest unit)</em>

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

<u>Formula for the Area of triangle</u>,  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.   <em>(Rounded to the nearest unit)</em>

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

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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}

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Step-by-step explanation:

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3 years ago
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Answer:

I assume that the function is:

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Now let's describe the general transformations that we need to use in this problem.

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g(x) = -f(x)

Reflection across the y-axis:

For a general function f(x), a reflection across the y-axis is written as:

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