How many distinct 9-letter words starting with “i” can be formed from the word “committee”?
1 answer:
We have already chosen "i" to be the first letter. This means that we still have 8 letters that we can use as the second letter. Once that letter is chosen, we still have 7 letters that we can use as the third letter, and so on. This means we have
ways of choosing the remaining 8 letters.
<u>However,</u> some of the 8 available letters are equal: we have 2 m's, 2 t's and 2 e's. This means that we overcounted the number of words on the reasoning above. We can correct the counting by dividing three times by 2!, one time for each of the repeated letters.
The final answer is
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Answer:
<A = 90°, <B = 71.6° <C = 18.4°
Step-by-step explanation:
Given the vertices of a △ABC to be A(−1, 6), B(2, 10), and C(7, −2)
Before we can get each angle of the triangle, we need to get its sides first.
Using the formula fie finding the distance between two points
D = √(x2-x1)²+(y2-y1)²
For side AB:
Given A(−1, 6), B(2, 10)
AB = √{2-(-1)}²+(10-6)²
AB =√3²+4²
AB = √25
AB = 5
For side AC:
Given A(-1,6) and C(7, −2)
AC = √{7-(-1)}²+(-2-6)²
AC = √8²+8²
AC = √128
AC = √64×2
AC = 8√2
For side BC:
Given B(2,10) and C(7, −2)
BC = √(7-2)²+(-2-10)²
BC = √5²+12²
BC = √25+144
BC = √169
BC = 13
First we need to get theta using cosine rule
|AC|² = |AB|+|BC| - 2|AB||AC| cos theta
(8√2)² = 5²+13²-2(5)(13)cos theta
128 = 169-130costheta
128-169 = -130costheta
-41= -130costheta
Costheta = -41/-130
Cos theta = 0.315
theta = arccos 0.315
theta = 71.64°
<B = 71.6°
Using sine rule to get <A,
BC/sin A = AC/sinB
13/sinA = 8√2/sin71.6°
8√2sinA = 13sin71.6°
SinA = 12.34/8√2
SinA = 1.09°
A = arcsin1
<A = 90.0°
<C = 180°-(71.6+90)
< C = 180-161.6
<C = 18.4°
Answer:
Width of the uniform path that surrounds the piece of land = 3 m
Step-by-step explanation:
Let the width of the path that surrounds the piece of land be x.
The situation described is sketched in the attached image to this solution.
The dimension of the Length of the piece of land including the uniform path that surrounds it = (40 + 2x) m
The Breadth of the piece of land including the uniform path that surrounds it = (25 + 2x) m
The area of the piece of land including the uniform path that surrounds it
= (Area of the piece of land) + (Area of the uniform path that surrounds it)
Area of the piece of land = Length × Breadth = 40 × 25 = 1000 m²
Area of the uniform path that surrounds it = 426 m² (Given in the question)
The area of the piece of land including the uniform path that surrounds it = 1000 + 426 = 1426 m²
But the area of the piece of land including the uniform path that surrounds it is also
= (Length of the piece of land including the uniform path that surrounds it) × (Breadth of the piece of land including the uniform path that surrounds it
= (40 + 2x) + (25 + 2x)
= 1000 + 80x + 50x + 4x²
= (4x² + 130x + 1000) m²
Equation these 2 areas
4x² + 130x + 1000 = 1426
4x² + 130x - 426 = 0
Solving the quadratic equation
x = 3 or -35.5
Since the width cannot be negative,
x = 3 m
Hope this Helps!!!
