Solution:
we have given the length of the rectangular card-
and, width-
now, the area of the card will be- length×width.
so, we have-
we also have given the length of ribbon-
and, width-
so, area of the ribbon is-
length×width
now the area of shaded gray region will be-
Hence, the required expression would the above expression.
Roots:
x-2 = 0
x = 2
x+3 = 0
x = -3
asymptotic:
none
lim (x-2)(x+3) = (inf -2)(inf+3 ) = inf
x ->inf
lim (x-2)(x+3) = (-inf-2)(-inf+3) = inf
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Answer:
The correct answer by solving question no 1 is
a=(d-c)/b
Step-by-step explanation:
ab+c=d
or,ab=d-c
Therefore,a=(d-c)/b
Hello :
<span>note :
an equation of the
circle Center at the w(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : </span><span>x²+y²-16x+6y+53=0
(</span>x²-16x) +( y²+6y ) +53 = 0
(x² -2(8)x +8² - 8²) +(y² +2(3)x -3²+3² ) +53=0
(x² -2(8)x +8²) - 8² +(y² +2(3)x +3²)-3² +53=0
(x-8)² +( y+3)² = 20
the center is : w(8,-3) and ridus : r = <span>√20</span>
Answer:
The value of MB = 8.4
Step-by-step explanation:
We know that the point of intersection of the Medians of a triangle is called the centroid of a triangle.
Thus,
For the given triangle ΔJKL,
- The point M is the centroid of the triangle.
We also know that the centroid is 2/3 of the distance from each vertex to the midpoint of the opposite side.
Also, each Median is split into two parts such that the longer part is 2 times the length of the smaller part.
In our case,
The median KB is split into two parts such that the longer part KM is 2 times the length of the smaller part MB.
i.e.
KM = 2 MB
Given KM = 16.8
so substitute KM = 16.8 in the equation KM = 2 MB
16.8 = 2 MB
MB = 16.8/2
MB = 8.4
Therefore, the value of MB = 8.4
