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2 years ago

Definitions of ASA SAS AAS, can someone dumb it down plz?

Mathematics

1 answer:

Veseljchak [2.6K]2 years ago

7 0

Answer:

Hope this helps.

Step-by-step explanation:

So for ASA, the two triangles have to have two angles congruent, and in the middle of those angles, they have to have a line that's congruent.

For SAS, the two triangles have to have two lines congruent, and in the middle of those lines, they have to have an angle that's congruent.

For AAS, the two triangles have to have two angles congruent, but the line that's congruent has to be on the side, not in the middle.

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What is the average rate of change of the function over the interval x = 0 to x = 6? f(x) = (2x - 1) / (3x + 5)

Sedaia [141]

Answer:

The average rate of change of the function over the interval x = 0 to x = 6 is $\frac{34}{195}$ .

Step-by-step explanation:

Geometrically speaking, the average rate of change ( $r$ ) of the function over a given interval is determined by equation of secant line, that is:

$r = \frac{f(b)-f(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds.

$f(a)$ , $f(b)$ - Function evaluated at lower and upper bounds.

If we know that $f(x) = \frac{2\cdot x -1 }{3\cdot x + 5}$ , $a = 0$ and $b = 6$ , then the average rate of change of the function over the interval is:

$f(a) = \frac{2\cdot (0)-1}{3\cdot (0) +5}$

$f(a) = -\frac{1}{5}$

$f(b) = \frac{2\cdot (6)-1}{3\cdot (6)+5}$

$f(b) = \frac{11}{13}$

$r = \frac{\frac{11}{13}-\left(-\frac{1}{5} \right) }{6-0}$

$r = \frac{34}{195}$

The average rate of change of the function over the interval x = 0 to x = 6 is $\frac{34}{195}$ .

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2 years ago

The diagonals of a parallelogram ABCD intersect at O.find ar(AOB),if ar(ABCD)is 44cm^2

prisoha [69]

Answer:

$11\text{ cm}^2$

Step-by-step explanation:

Given: In a parallelogram ABCD, diagonals intersect at O and ar(ABCD) is $44 \text{cm}^2$ .

We need to find the area of triangle AOB.

We know that each diagonal divide the parallelogram in two equal parts and diagonals bisect each other.

It means both diagonals divide the parallelogram in 4 equal parts.

$ar(AOB)=\dfrac{1}{4}\times ar(ABCD)$

$ar(AOB)=\dfrac{1}{4}\times 44$

$ar(AOB)=11$

Hence, the values of ar(AOB) is $11\text{ cm}^2$ .

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3 years ago

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Step-by-step explanation:

y = 3cos(x) - 5

Domain = 2sin(x)

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Answer:

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I hope this is good enough:

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