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Answer:

#x^3+3x^2-1#

Explanation:

#"one way is to use the divisor as a factor in the numerator"#

#"consider the numerator"#

#color(red)(x^3)(x+2)color(magenta)(-2x^3)+5x^3+6x^2-x-2#

#=color(red)(x^3)(x+2)color(red)(+3x^2)(x+2)cancel(color(magenta)(-6x^2))cancel(+6x^2)-x-2#

#=color(red)(x^3)(x+2)color(red)(+3x^2)(x+2)color(red)(-1)(x+2)cancel(color(magenta)(+2))cancel(-2)#

#"quotient "=color(red)(x^3+3x^2-1)," remainder "=0#

#rArr(x^4+5x^3+6x^2-x-2)/(x+2)#

#=(cancel((x+2))(x^3+3x^2-1))/cancel((x+2))=x^3+3x^2-1#

Answer:

One should complete the squares so that the equation may be written in one of the two following forms:

#(x-h)^2/a^2+(y-k)^2/b^2=1" [1]"#

#(y-k)^2/a^2+(x-h)^2/b^2=1" [2]"#

where #a>b#

Explanation:

Given:

#x^2 + 8y^2 − 8x − 16y − 40 = 0#

Add 40 to both sides:

#x^2 + 8y^2 − 8x − 16y = 40#

Group the x terms and the y terms together:

#(x^2 − 8x) + (8y^2 − 16y) = 40#

We cannot complete the square unless the leading coefficient is 1, therefore, we remover a factor of 8 from the y terms:

#(x^2 − 8x) + 8(y^2 − 2y) = 40#

Because #(x-h)^2= x^2-2hx+h^2# we want to insert an #h^2# into the x term's group but we must, also add #h^2# to the right side so that equality is maintained:

#(x^2 − 8x+h^2) + 8(y^2 − 2y) = 40+h^2#

Matching the x terms with the general pattern, #(x-h)^2= x^2-2hx+h^2#, we observe that the equation

#-2hx = -8x#

will allow us to solve for the value of h:

#h = 4#

This means that #h^2# on the right side becomes 16 and the group of x terms become #(x-4)^2#

#(x − 4)^2 + 8(y^2 − 2y) = 40+16#

Combine like terms:

#(x − 4)^2 + 8(y^2 − 2y) = 56#

We want insert #k^2# into the y terms but to maintain equality we must add #8k^2# to the right side:

#(x − 4)^2 + 8(y^2 − 2y+ k^2) = 56+8k^2#

Matching the y terms with the general pattern, #(y-k)^2= y^2-2ky+k^2#, we observe that the equation

#-2ky = -2y#

will allow us to solve for the value of k:

#k = 1#

This means that #8k^2# on the right side becomes 8 and the group of y terms become #(y-1)^2#

#(x − 4)^2 + 8(y − 1)^2 = 56+8#

Combine like terms:

#(x − 4)^2 + 8(y − 1)^2 = 64#

Divide both sides by 64:

#(x − 4)^2/64 + (y − 1)^2/8 = 1#

Write the denominators as squares:

#(x − 4)^2/8^2 + (y − 1)^2/(2sqrt2)^2 = 1#

This is the same form as equation [1],

#(x-h)^2/a^2+(y-k)^2/b^2=1" [1]"#

The center, #(h,k) = (4,1)#
The vertices are, #(h-a,k) = (-4,1)# and #(h+a,k)= (12,1)#
The foci are, #(h-sqrt(a^2-b^2),k) = (4-sqrt56,1)# and #(h+sqrt(a^2-b^2),k) = (4+sqrt56,1)#
The eccentricity is #sqrt(a^2-b^2)/a = sqrt56/8#

Answer:

#(-2,5)" is the maximum"#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#• " if "a>0" then f(x) is a minimum"#

#• " if "a<0" then f(x) is a maximum"#

#(h,k)" will be the max/min turning point of f(x)"#

#"to express "f(x)=-1/2x^2-2x+3" in vertex form"#

#"use the method of "color(blue)"completing the square"#

#• " the coefficient of the "x^2" term must be 1"#

#rArrf(x)=-1/2(x^2+4x)+3#

#• " add/subtract "(1/2"coefficient of x-term")^2" to"#
#x^2+4x#

#f(x)=-1/2(x^2+2(2)xcolor(red)(+4)color(red)(-4))+3#

#color(white)(f(x))=-1/2(x+2)^2+2+3#

#color(white)(f(x))=-1/2(x+2)^2+5larrcolor(blue)"in vertex form"#

#rArrcolor(magenta)"vertex "=(-2,5)" and "a<0#

#rArr"maximum turning point "=(-2,5)#
graph{(y+1/2x^2+2x-3)((x+2)^2+(y-5)^2-0.04)=0 [-10, 10, -5, 5]}

Answer:

The domain is #x in (-oo,-2)uu(-2,2)uu(2,+oo)#.
The range is #y in RR#

Explanation:

The function is #f(x)=x/(x^2-4)=x/((x+2)(x-2))#

As the denominator must be #!=0#, so

#x!=2# and #x!=-2#

The domain is #x in (-oo,-2)uu(-2,2)uu(2,+oo)#

To find the range, proceed as follows :

#y=x/(x^2-4)#

#y(x^2-4)=x#

#yx^2-4y=x#

#yx^2-x-4y=0#

In order for this quadratic equation in #x# to have solutions, the discriminant #Delta>=0#

#b^2-4ac=(-1)^2-4(y)(-4y)>=0#

#1+16y^2>=0#

Therefore,

#AA y in RR,#, #Delta>=0#

The range is #y in RR#

graph{x/(x^2-4) [-10, 10, -5, 5]}

Answer:

#x^2/9-y^2/16=1#.

Explanation:

Recall that, for the Hyperbola # S : x^2/a^2-y^2/b^2=1#, the

asymptotes are given by, #y=+-b/a*x;" where, "a,b gt 0#.

In our Case, the asymptotes are, #y=+-4/3*x#.

#:. b/a=4/3...........................................................................(1)#.

Next, #(3sqrt2,4) in S rArr (3sqrt2)^2/a^2-4^2/b^2=1#.

#:. 18/a^2-16/b^2=1#.

Multiplying by #b^2#, we get, #18*b^2/a^2-16=b^2#.

By #(1)#, then, #18*(4/3)^2-16=b^2, i.e., b^2=16#.

#:. b=4 (because, b >0), and, (1) rArr a=3, or, a^2=9#.

With #a^2=9, and b^2=16#, the reqd. eqn. of the Hyperbola is,

#x^2/9-y^2/16=1#.

Answer:

#r=5#, #s=3# and #t=7#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,1,1,|,15),(1,0,1,|,12),(0,1,1,|,10))#

I have written the equations not in the sequence as in the question in order to get #1# as pivot.

Perform the folowing operations on the rows of the matrix

#R2larrR2-R1#

#A=((1,1,1,|,15),(0,-1,-0,|,-3),(0,1,1,|,10))#

#R1larrR1+R2#; #R3larrR3+R2#

#A=((1,0,1,|,12),(0,-1,-0,|,-3),(0,0,1,|,7))#

#R1larrR1-R3#

#A=((1,0,0,|,5),(0,-1,-0,|,-3),(0,0,1,|,7))#

#R2larr(R2)*(-1)#

#A=((1,0,0,|,5),(0,1,0,|,3),(0,0,1,|,7))#

Thus #r=5#, #s=3# and #t=7#

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