Answer:
65625/4(x^5)(y²)
Step-by-step explanation:
Using binomial expansion
Formula: (n k) (a^k)(b ^(n-k))
Where (n k) represents n combination of k (nCk)
From the question k = 5 (i.e. 5th term)
n = 7 (power of expression)
a = 5x
b = -y/2
....................
Solving nCk
n = 7
k = 5
nCk = 7C5
= 7!/(5!2!) ------ Expand Expression
=7 * 6 * 5! /(5! * 2*1)
= 7*6/2
= 21 ------
.........................
Solving (a^k) (b^(n-k))
a = 5x
b = -y/2
k = 5
n = 7
Substituting these values in the expression
(5x)^5 * (-y/2)^(7-5)
= (3125x^5) * (-y/2)²
= 3125x^5 * y²/4
= (3125x^5)(y²)/4
------------------------------------
Multiplying the two expression above
21 * (3125x^5)(y²)/4
= 65625/4(x^5)(y²)
Answer: Each dozen of chocolate chip cookies costs $1.50/dozen. So if you sell C dozens of them, you win C*$1.50 dollars, where C is a positive integer number. Similar case for the lemon ones, the baker will win L*$1.00 dollars if he sells L dozens of them ( where L is a positive integer), the total charge is the sum of both parts; T = L*$1.00 + C*$1.50.
Answer:
Step-by-step explanation:
The first step is to determine the areas of the inner and the bigger circles.
The formula for determining the area of a circle is expressed as
Area = πr²
Where
r represents the radius of the circle.
π is a constant whose value is 3.14
Considering the inner circle,
diameter = 148.4 cm
Radius = 3 inches
Area = 3.14 × 3² = 28.26 inches²
Considering the bigger circle,
Radius = 7 inches
Area = 3.14 × 7² = 153.86 inches²
The area between the bigger and inner circles is
153.86 - 28.26 = 125.6 inches²
Probability = number of favourable outcome/number of total outcome.
the probability you land inside the bigger circle but outside smaller circle is
125.6/153.86 = 0.82
Segment NO is parallel to the segment KL.
Solution:
Given KLM is a triangle.
MN = NK and MO = OL
It clearly shows that NO is the mid-segment of ΔKLM.
By mid-segment theorem,
<em>The segment connecting two points of the triangle is parallel to the third side and is half of that side.</em>
⇒ NO || KL and 
Therefore segment NO is parallel to the segment KL.
