Answer:
NO
Step-by-step explanation:
Simply enter 5, 10 in the x and y boxes to get the answer.
10=6(5)-18
10=30-18
10=12
As a consequence, it isn't a feasible option.
In this problem, we can imagine that all the points
connect to form a triangle. The three point or vertices are located on the
pitcher mount, the home plate and where the outfielder catches the ball. So in
this case we are given two sides of the triangle and the angle in between the
two sides.
<span>With the following conditions, we can use the cosine law
to solve for the unknown 3rd side. The formula is:</span>
c^2 = a^2 + b^2 – 2 a b cos θ
Where,
a = 60.5 ft
b = 195 ft
θ = 32°
Substituting the given values:
c^2 = (60.5)^2 + (195)^2 – 2 (60.5) (195) cos 32
c^2 = 3660.25 + 38025 – 20009.7
c^2 = 21,675.56
c = 147.23 ft
<span>Therefore the outfielder throws the ball at a distance of
147.23 ft towards the home plate.</span>
Answer:
Length = 16 inches and width = 12inches
Step-by-step explanation:
A = 192 in² . The width is 4 + 1/2x. therefore let x = length
Area = LW
192 = x(4 + 1/2 x)
192 = 4x + 1/2x²
384 = 8x + x² Multiply each term by 2 to make a
384 + 16 = x² + 8x + 16 Complete the square
400 = (x + 4)²
± 20 = x + 4
20 = x + 4 or -20 = x + 4
20 - 4 = x or -20 - 4 = x
16 = x or -24 = x reject the negative amount
Length = 16
width = 4 + 1/2(16); 4 + 8 = 12
X
+
6
−
5
=
8
−
2
+
6
−
5
=
8
−
2
Solve
1
Subtract the numbers
x
+
6
−
5
=
8
−
2
+
6
−
5
=
8
−
2
x
+
1
=
8
−
2
+
1
=
8
−
2
2
Subtract the numbers
x
+
1
=
8
−
2
+
1
=
8
−
2
x
+
1
=
6
+
1
=
6
3
Subtract
1
1
from both sides
x
+
1
=
6
+
1
=
6
x
+
1
−
1
=
6
−
1
+
1
−
1
=
6
−
1
4
Simplify
Subtract the numbers
Subtract the numbers
x
=
5
=
5
We know that
<span>the regular hexagon can be divided into 6 equilateral triangles
</span>
area of one equilateral triangle=s²*√3/4
for s=3 in
area of one equilateral triangle=9*√3/4 in²
area of a circle=pi*r²
in this problem the radius is equal to the side of a regular hexagon
r=3 in
area of the circle=pi*3²-----> 9*pi in²
we divide that area into 6 equal parts------> 9*pi/6----> 3*pi/2 in²
the area of a segment formed by a side of the hexagon and the circle is equal to <span>1/6 of the area of the circle minus the area of 1 equilateral triangle
</span>so
[ (3/2)*pi in²-(9/4)*√3 in²]
the answer is
[ (3/2)*pi in²-(9/4)*√3 in²]
