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Answer:

#theta ~~ 102.7°#

Explanation:

Compute vector C:

#barC=barA -barB = (7-4)hati + (-5 - -8)hatj + (6 - 9)hatk#

#barC = 3hati + 3hatj - 3hatk#

Compute the dot-product of vectors A and C:

#barA*barC = (7)(3) + (-5)(3) + (6)(-3)#

#barA*barC = (7)(3) + (-5)(3) + (6)(-3)#

#barA*barC = -12#

There is another form for the equation of the dot-product that contains the angle between the two vectors:

#barA*barC = |barA||barC|cos(theta) = -12#

Compute the magnitude of vector A:

#|barA| = sqrt(7^2 + (-5)^2 + 6^2)#

#|barA| = sqrt(110)#

Compute the magnitude of vector C:

#|barC| = sqrt(3^2 + 3^2 + (-3)^2)#

#|barC| = 3sqrt(3)#

#sqrt(110)(3sqrt(3))cos(theta) = -12#

#cos(theta) = -12/(sqrt(110)(3sqrt(3))#

#theta = cos^-1(-4/(sqrt(330)))#

#theta ~~ 102.7°#

Answer:

(see below)

Explanation:

Based on the Pythagorean Theorem, we know that
#color(white)("XXX")a^2+b^2=c^2

#rarr b=sqrt(c^2-a^2)=sqrt(6^2-5^2)=sqrt(11)#
enter image source here

#sin = "opposite"/"hypotenuse"color(white)("XX")rarr sin(A)=5/6#

#csc = "hypotenuse"/"opposite" color(white)("XX")rarr csc(A)=6/5#

#cos = "adjacent"/"hypotenuse"color(white)("XX")rarr cos(A)=sqrt(11)/6#

#sec = "hypotenuse"/"adjacent" color(white)("XX")rarr sec(A)=6/sqrt(11)#

#tan = "opposite"/"adjacent"color(white)("XX")rarr tan(A)=5/sqrt(11)#

#cot = "adjacent"/"opposite" color(white)("XX")rarr cot(A)=sqrt(11)/5#

Answer:

#sec(cos^(-1)(1/2)) = 2#

Explanation:

Another way, without calculating #cos^(-1)(1/2)#:

Consider a right triangle with an angle #theta = cos^(-1)(1/2)#. Then #cos(theta) = 1/2#, meaning the ratio of its adjacent side to the hypotenuse is #1/2#. Thus the ratio of the hypotenuse to its adjacent side, that is, #sec(theta)#, is #2/1 = 2#.

Thus #sec(cos^(-1)(1/2)) = sec(theta) = 2#

Note that this same reasoning shows that in general, #sec(cos^(-1)(x)) = 1/x#

Given expression #=arcsin((x+1)/sqrt(2*(x²+1)))#
Let #x = tantheta#

So #theta=tan^-1x#

Inserting #x = tantheta# the given expression becomes

#=arcsin((tantheta+1)/sqrt(2*(tan^2theta+1)))#

#=arcsin((sintheta/costheta+1)/sqrt(2*(sec^2theta)))#

#=arcsin(1/sqrt2((sinthetasectheta+1)/(sectheta)))#

#=arcsin(1/sqrt2((sinthetacancelsectheta)/cancelsectheta+1/(sectheta)))#

#=arcsin(1/sqrt2(sintheta+costheta))#

#=arcsin(1/sqrt2sintheta+1/sqrt2costheta)#

#=arcsin(cos(pi/4)sintheta+sin(pi/4)costheta)#

#=arcsin(sin(theta+pi/4))#

#=theta+pi/4#

#=tan^-1x+pi/4#

Answer:

Set your compass to a radius of 6, put the center point at #(6, 0)#, and draw a circle.

Explanation:

Multiply both sides of the equation by r:

#r^2 = 12rcos(theta)#

Substitute #x^2 + y^2# for #r^2# and #x# for #rcos(theta)#

#x^2 + y^2 = 12x#

The standard form of this type of equation (a circle) is:

#(x - h)^2 + (y - k)^2 = r^2#

To put the equation in this form, we need to complete the square for both the x and y terms. The y term is easy, we merely subtract #0# in the square:

#x^2 + (y - 0)^2 = 12x#

To complete the square for the x terms, we add #-12x + h^2# both sides of the equation:

#x^2 - 12x + h^2 + (y - 0)^2 = h^2#

We can use the pattern, #(x - h)^2 = x^2 - 2hx + h^2# to find the value of h:

#x^2 - 2hx + h^2 = x^2 - 12x + h^2#

#-2hx = -12x#

#h = 6#

Substitute 6 for h into the equation of the circle:

#x^2 - 12x + 6^2 + (y - 0)^2 = 6^2#

We know that the first three terms are a perfect square with h = 6:

#(x - 6)^2 + (y - 0)^2 = 6^2#

This is a circle with a radius of 6 and a center point #(6, 0)#

Answer:

Please see the explanation.

Explanation:

Here is the graph:

Desmos.com

This looks like a cosine function that has been multiplied by an amplitude and phase shift to the right:

#f(x) = Acos(x - phi) = sin(x) + cos(x)#

My graphing tool allows me to obtain values of points and I can tell you that #A = sqrt(2)#

#cos(x - phi) = 1/sqrt(2)sin(x) + 1/sqrt(2)cos(x)#

This fits the trigonometric identity:

#cos(x - phi) = sin(x)sin(phi) + cos(x)cos(phi)#

where #cos(phi) = sin(phi) = 1/sqrt(2)#

This happens at #phi = pi/4#

The graphing tool confirms that the x coordinates are shift by #pi/4#.

#f(x) = sqrt(2)cos(x - pi/4)#

Substitute #f^-1(x)# for every x:

#f(f^-1(x)) = sqrt(2)cos(f^-1(x) - pi/4)#

The left side becomes x by definition:

#x = sqrt(2)cos(f^-1(x) - pi/4)#

Make the #sqrt(2)# on the right disappear by multiplying both sides by #sqrt(2)/2#:

#sqrt(2)/2x = cos(f^-1(x) - pi/4)#

Use the inverse cosine on both sides:

#cos^-1(sqrt(2)/2x) = f^-1(x) - pi/4#

Solve for #f^-1(x)#:

#f^-1(x) = cos^-1(sqrt(2)/2x) + pi/4#

To confirm that this is truly an inverse, verify that #f(f^-1(x)) = f^-1(f(x)) = x#

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