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3 years ago

The law of cosines is a² + b² - 2abcosC = c². find the value of 2abcosC

Mathematics

2 answers:

taurus [48]3 years ago

7 0

$\bf a^2+b^2-(2ab)cos(C)=c^2\implies a^2+b^2-c^2=(2ab)cos(C)$

Elina [12.6K]3 years ago

3 0

Answer:

We need a diagram please

Step-by-step explanation:

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find three consecutive even integers so that the twice the sum of the second and third is twelve less than six times the first

zysi [14]

n, n + 2, n + 4 - three consecutive even integers

the twice the sum of the second and third: 2[(n + 2) + (n + 4)]

twelve less than six times the first: 6n - 12

The equation:

2[(n + 2) + (n + 4)] = 6n - 12

2(n + 2 + n + 4) = 6n - 12

2(2n + 6) = 6n - 12 use distributive property

(2)(2n) + (2)(6) = 6n - 12

4n + 12 = 6n - 12 subtract 12 from both sides

4n = 6n - 24 subtract 6n from both sides

-2n = -24 divide both sides by (-2)

n = 12

n + 2 = 12 + 2 = 14

n + 4 = 12 + 4 = 16

<h3>Answer: 12, 14, 16</h3>

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3 years ago

Solve the proportion <img src="https://tex.z-dn.net/?f=%5Cfrac%7BX%2B1%7D%7B3%7D" id="TexFormula1" title="\frac{X+1}{3}" alt="\f

Ahat [919]

D.) 2 is the answer

i think

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3 years ago

Differentiate 6+4x - x^2

vfiekz [6]

Answer:

2x+4

Step-by-step explanation:

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2 years ago

An employee working in a machine shop is exposed to three different sources which emit noises at 81 dB, 91 dB, and 88 dB. What i

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4 years ago

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |subtract $y$ from both sides

$\dfrac{2}{3}x-y+b=0$ |multiply both sides by 3

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |multiply both sides by $\sqrt{13}$

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |divide both sides by 3

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

4 0

3 years ago