<img src="https://tex.z-dn.net/?f=c%5E2%3Da%5E2%2Bb%5E2-2ab%20cosC" id="TexFormula1" title="c^2=a^2+b^2-2ab cosC" alt="c^2=a^2+b

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nika2105 [10]

2 years ago

^2-2ab cosC" align="absmiddle" class="latex-formula">
This is one form of the Law of?

Mathematics

1 answer:

Ad libitum [116K]2 years ago

8 0

$\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies 
c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\ \quad \\

\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}=cos(C)\implies cos^{-1}\left(\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}\right)=\measuredangle C$

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What are the solutions of the system y=x^2+3x-4 y=2x+2

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3 years ago

Which graph has a slope of 4/5

Lana71 [14]

Answer: whichever graph that goes up 4 and over 5

Step-by-step explanation:

rise over run, 4 up, 5 right

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3 years ago

In triangle ABC, line A.B.=CB Angle ABC equals 4X -3, and angle CAB equals X -3, what is angle ACB

Soloha48 [4]

Answer:

∠ACB = 28.5

Step-by-step explanation:

In triangle ABC,

angle CAB = x-3

angle ABC = 4X-3

It is also given that AB=CB

We know that if two sides of any triangle are equal then corresponding sides must also be equal.

Hence

angle ACB = angle CAB

angle ACB = x-3 {Given that angle CAB = x-3 }

We know that sum of all three angles of a triangle is 180 degree so let's add given angles

angle ACB + angle CAB + angle ABC = 180

(x-3) + (x-3) + (4x-3) = 180

x-3 + x-3 + 4x-3 = 180

6x-9 = 180

6x = 180+9

6x = 189

x = 189/9

x= 31.5

Now plug value of x into

angle ACB = x-3 = 31.5-3 = 28.5

Hence final answer is

∠ACB = 28.5

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2 years ago

Cassie rolls a fair number cube with 6 faces labeled 1 through 6. She rolls the number cube 300 times.Which result is most likel

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3 years ago

Express the function graphed on the axes below as a piecewise function.

adoni [48]

Answer:

f(x) = {x-3 for x ≤ -1; -3x+14 for x > 5}

Step-by-step explanation:

To write the piecewise function, we can consider the pieces one at a time. For each, we need to define the domain, and the functional relation.

<h3>Left Piece</h3>

The domain is the horizontal extent. It is shown as -∞ to -1, with -1 included.

The relation has a slope (rise/run) of +1, and would intersect the y-axis at -3 if it were extended.

The first piece can be written ...

f(x) = x-3 for x ≤ -1

<h3>Right Piece</h3>

The domain is shown as 5 to ∞, with 5<em> not included</em>.

The relation is shown as having a slope (rise/run) of (-3)/(1) = -3. If extended, it would intersect the point (5, -1), so we can write the point-slope equation as ...

y -(-1) = -3(x -5)

y = -3x +15 -1 = -3x +14

The second piece can be written ...

f(x) = -3x +14 for x > 5

<h3>Whole function</h3>

Putting these pieces together, we have ...

$\boxed{f(x)=\begin{cases}x-3&\text{for }x\le-1\\-3x+14&\text{for }5 < x\end{cases}}$

_____

<em>Additional comment</em>

Sometimes it is convenient to write inequalities in number-line order (using < or ≤ symbols). This gives a visual indication of where the variable stands in relation to the limit(s). Perhaps a more conventional way to write the domain for the second piece is, <em>x > 5</em>.

3 0

2 years ago