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Answer:

New coordinates of #color(red)(A ((1),(1))#

Change (reduction) in distance from #bar(AB), bar(A'B)# is

#=> sqrt13 - 3 ~~ color(green)(0.61)#

Explanation:

A (1, -1), B (-2,1) and rotated by #(3pi)/2# clockwise about the origin.

Point A moves from IV quadrant to I quadrant.

#A((1),(-1)) ->A'((1),(1))#

Distance #vec(AB) = sqrt((1+2)^2 + (-1-1)^2) = color(blue)(sqrt13#

Distance #vec(A'B) = sqrt((1+2)^2 + (1-1)^2) = color(blue)3#

Change (reduction) in distance from #bar(AB), bar(A'B)# is

#=> sqrt13 - 3 ~~ color(green)(0.61)#

Answer:

#A_2 = (A_1/2) (1 + (b / (a+b)))#

Explanation:

Area of a trapizoin #A = ((a+b) / 2) * h# where a, b are the parallel sides and h the distance between them.

#A_1 = ((a+b)/2) * h# Eqn (1)

Now base a has been halved or the new base is a/2
#A_2 = (((a/2) + b)/2) * h#

#A_2 = ((a+2b)/4)*h# Eqn (2)

Dividing Eqn(2) by (1),

#A_2 / A_1 = (((a+2b)/cancel4^2) * cancelh) / (((a+b)/cancel2) * cancelh)#

#A_2 = ((a + 2b) /(2* (a + b))) * A_1#

#A_2 = (A_1/2) (1 + (b / (a+b)))#

Answer:

Case 1 : ED is the altitude (perpendicular to GH) #x = color(brown)(78.92# ft

Case 2 : ED is the median #x = color(blue)(60# ft

Case 3 :ED is the angular bisector of #GhatEH# #x = color(green)(68# ft

Explanation:

Assuming enter image source here

Since it has not been given, ED is perpendicular to GH or D is the mid point of GH or ED is the angular bisector of #GhatEH#

Let's consider all these three cases.

Case 1 : ED is the altitude (perpendicular to GH)

In right triangle DEG,

#DE^2 = GE62 - DG^2 = 99.2^2 - 62^2# Eqn (1)

In right triangle DEH,

#DE^2 = EH^2 - DH^2 = 112^2 - (x+2)^2# Eqn (2)

Comparing Eqns (1), (2),

#112^2 - (x+2)^2 = 99.2^2 - 62^2#

#(x+2)^2 = 112^2 - 99.2^2 + 62^2 = 6547.36#

#(x+2) ~~ 80.92#

#x = 80.92 - 2 = color(brown)(78.92# ft

Case 2 : ED is the median

i.e DH = DG #

#x + 2 = 62# or #x = 62 - 2 = color(blue)(60# ft

Case 3 : ED is the angular bisector of #GhatEH

As per angular bisector theorem,

#(EG) / (EH) = (DG) / (GH)#

#99.2 / 112 = 62 / (x + 2)#

#(x +2) = (62 * 112) / 99.2 = 70#

#x = 70 - 2 = color(green)(68)# ft

Answer:

#x^2+y^2-6x+2y-28=0#

Explanation:

The length of perpendicular from #(3,-1)# to #2x-5y+18=0# is

#|(2xx3-5xx(-1)+18)/sqrt(2^2+5^2)|=29/sqrt29=sqrt29#

Now as circle cuts off an intercept of length #6#, it intercepts at a distance of #3# on each side from the foot of the perpendicular.

Hence radius of circle is #sqrt((sqrt29)^2+3^2)=sqrt38#

and equation of circle is

#(x-3)^2+(y+1)^2=38#

or #x^2+y^2-6x+2y-28=0#

graph{(x^2+y^2-6x+2y-28)(2x-5y+18)((x-3)^2+(y+1)^2-0.03)=0 [-9.75, 10.25, -4.04, 6.38]}

Answer:

#"arc DF "=pi~~3.14" feet to 2 dec. places"#

Explanation:

#"the length of the arc is calculated using"#

#• " length of arc "="circumference "xx"fraction of circle"#

#color(white)(xxxxxxxxxxxx)=2pirxx60/360#

#color(white)(xxxxxxxxxxxx)=6pixxcancel(60)^1/cancel(360)^6#

#color(white)(xxxxxxxxxxxx)=(cancel(6) pi)/cancel(6)=pi~~3.14#

Answer:

See below.

Explanation:

Here we are formulating an equation to solve for #x#.

We know that the interior angles of any triangle adds up #180# degrees.

We have three angles given:

#60#
#x#
#3x#

This means that :

#60+3x+x=180#

Now we collect like terms to simplify.

#60+4x=180#

Now we solve like any linear equation by isolating the variable on one side of the equation with the constant on the other.

Here we must subtract #60# from both sides to isolate the #x#.

#therefore 60 + 4x -60 = 180 -60#

#=>4x=120#
We want one #x#, therefore we divide by the coefficient of #x# on both sides.

Here we divide by #4#

#4x=120#
#=>x=30#

We can check if we are right by putting our value of #x# back into our formulated equation above.

#60 + (4*30) = 60+120 = 180#

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  • Gió answered · 17 hours ago
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