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2 years ago

The law of cosines is a^2+b^2-2abcosC=c^2. Find the value of 2abcosC.

Mathematics

2 answers:

Ghella [55]2 years ago

5 0

Answer:

(D) $7=2abcosC$

Step-by-step explanation:

It is given from figure that c=1, a=2 and b=2, thus using the law of cosines, we get

$c^2=a^2+b^2-2abcosC$

Substituting the given values, we get

$(1)^2=(2)^2+(2)^2-2abcosC$

$1=4+4-2abcosC$

$1=8-2abcosC$

$1-8=-2abcosC$

$-7=-2abcosc$

$7=2abcosC$

Thus, the value of $2abcosC$ is $7$ .

Hence, option (D) is correct.

polet [3.4K]2 years ago

4 0

2ab x cos (C) = 7 because

a^2 + b^2 - 2ab x cos(C) = c^2
2ab x cos(C) = a^2 + b^2 - c^2
2ab x cos(C) = 2^2 + 2^2 - 1^2
2ab x cos (C) = 7

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F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

34kurt

Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

$g(f(x)) = 3x^4 + 18x^2 + 18$

Checking for inverse functions

$f(x) = 3x^2 + 9$

Represent f(x) with y

$y = 3x^2 + 9$

Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

Note that the resulting value of g(x) is not the same as $g(x) = \dfrac{1}{3}x^2 - 9$

<em>Hence, f(x) and g(x) and not inverse functions</em>

4 0

2 years ago

Zach walked his dog for 7/8 hour. They spent 1/2 of the time at the beach. How much time did Zach and his dog spend at the beach

Tresset [83]

Answer: $\frac{7}{16}hour$

Step-by-step explanation:

You know that the total time was: 7/8 hour

You also know that they spent 1/2 of the time at the beach.

Therefore, to calculate the time Zach and his dog spend at the beach (which you can call x) , you must multiply both fractions (7/8 and 1/2).

Then, you obtain the following result:

$x=\frac{7}{8}hour*\frac{1}{2}\\\\x=\frac{7}{16}hour$

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2 years ago

Let x = 8.99999 . (a) Is x < 9 or is x = 9? x < 9 It is neither; x > 9 x = 9 x < 9 and x = 9 It cannot be determined

qaws [65]

Answer: idk

Step-by-step explanation:

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2 years ago

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Answer:

30-100/1000*120

Step-by-step explanation:

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2 years ago

The equations 3 x minus 4 y = negative 2, 4 x minus y = 4, 3 x + 4 y = 2, and 4 x + y = negative 4 are shown on the graph below.

stich3 [128]

Answer:

(–1.4, 1.5)

Step-by-step explanation:

The blue line and the purple line are the lines corresponding to the equations of interest. Their point of intersection is in the 2nd quadrant, so is nearest to ...

(–1.4, 1.5)

It can be useful to understand that for equations in standard form:

ax +by = c

the x- and y-intercepts are ...

x-intercept: c/a . . . . value of x for y = 0
y-intercept: c/b . . . . value of y for x = 0

For the equations of interest, the first has intercepts of ...

x=2/3, y=1/2 . . . . graphed line makes a 1st-quadrant triangle with the axes (blue line)

And the second has intercepts of ...

x=-1, y=-4 . . . . graphed line makes a 3rd-quadrant triangle with the axes (purple line)

Since the purple line has a steeper slope, the point of intersection of the lines will be in the 2nd quadrant. There is only one 2nd-quadrant answer choice: (-1.4, 1.5).

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2 years ago

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