TRUE OR FALSE."Some isosceles triangles are not equilateral.''

1 answer:

8 0

This is true. An isosceles triangle is one that has two equal sides and two equal angles. The third side can be any length and the third angle will also vary depending on the other two angles. An equilateral triangle is one where all sides and angles are equal. An isosceles can be equilateral, but it does not have to be.

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Hi can someone pls help me with this!!!

Eva8 [605]

the domain is the x value (first number) and the range is the y value (second number)

(if a number appears more than once in the domain or range, like in number 1 you don't have to write it again.)

to graph the domain and range you just plot the points,

and to map them you have to put the x values in the first oval and the y values in the second, usually in order from smallest to largest.

then you have to draw arrows connecting each x value to the y value that was in the same pair. just like when writing down the domain/range, if a number comes up again you don't have to write it down again. instead, you might have two or more arrows connecting to the same number.

8 0

3 years ago

Find the area enclosed between f(x)=0.3x2+7 and g(x)=x from x=−4 to x=8.

denis23 [38]

First, we sketch a picture to get a sense of the problem. g(x)=x is a diagonal line through (0,0) with slope = = 1. Since we are interested in the area between x = -4 and x = 8, we find the points on the line at these values. These are (-4, -4) and (8,8).

f(x) is a parabola. It's lowest point occurs when x = 0. It is the point (0,7). At x = -4 and x=8 it has the values 11.8 and 26.2 respectively. That is, it contains the points (-4, 11.8) and (8,26.2).

From these we make a rough sketch (see attachment). This is a sketch and mine is very incorrect when it comes to scale but what matters here is which of the curves is on top, which is below and whether they intersect anywhere in the interval, so my rough sketch is good enough. From the sketch we see that f(x) is always above (greater than) g(x).

To find the area between the curves over the given interval we integrate their difference and since f(x) is strictly greater than g(x) we subtract as follows: f(x) - g(x). The limits of integration are the values -4 and 8 (the x-values between which we are looking for the area.

Now let's integrate:
$\int\limits^{8}_ {-4}f(x)-g(x) \, dx = \int\limits^{8}_ {-4}.3 x^{2} +7-x \, dx$
The integral yields: $[tex](\frac{.3 (8)^{3} }{3} +7(8)- \frac{ (8)^{2} }{2}) -(\frac{.3 (-4)^{3} }{3} +7(-4)- \frac{ (-4)^{2} }{2}) = 117.6$ [/tex]
We evaluate this for 8 and for -4 subtracting the second FROM the first to get: