Answer:
The perimeter of parallelogram PQRS = 
⇒
2nd answer
Step-by-step explanation:
* <em>Lets revise some properties of the parallelogram</em>
- Each two opposite sides are parallel
- Each two opposite sides are equal
- Its <em>perimeter</em> is the <em>twice the sum of two adjacent sides</em>
* <em>Lets solve the problem</em>
∵ PQRS is a parallelogram
∵ The length of side SR is 
∵ The length of side QR is 
∵ <em>SR and RQ are two adjacent sides</em>
∵ The perimeter of parallelogram PQRS = 2(RQ + SR)
∴ The perimeter of parallelogram PQRS = 
∵ 
= 
∵ 
= 
∴ The perimeter of parallelogram PQRS =
Answer:
Closed circle going right
Step-by-step explanation:
Answer:
a) The easiest way is to use a calculator such as a Casio brand or Sharp brand and key in 9C7 to get 36.
However, you can also use this formula: 
where n is the total number of items and r is how many you are selecting.
Plug in values:
b) The easiest way is to use a calculator such as a Casio brand or Sharp brand and key in 9P7 to get 181440.
However, you can also use this formula: 
where n is the total number of items and r is how many you are selecting.
Plug in values:
There is a formula for parabola
y-k=1/4p(x-h)²
and the focal length is p
So the answer is 2/3
9514 1404 393
Answer:
C and D
Step-by-step explanation:
Extraneous solutions arise in a couple of cases, both of which are illustrated in this list of answer choices.
1. Square roots (and other even-index radicals). The square root is defined only for positive arguments. When the radical is removed by squaring the equation, the possibility of negative arguments is introduced. Hence, there may be extraneous solutions.
<u>Example</u>:
x = √(x +2) ⇒ x² = x+2 . . . has 2 roots, one of which is extraneous
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2. Rational function equations. Such an equation is only defined when the denominator is not zero. A method often taught for solution of these equations is "cross-multiplying"--essentially multiplying by the product of the denominators. In the case where this product is not the least common denominator, the equation will have an extra factor that can introduce an extraneous root.
__
The answer choices C and D are likely to have extraneous roots based on the above observations.
_____
<em>Additional comment</em>
Choice D would usually be solved by cross-multiplying. That would give the quadratic equation (x -3)² = 0, which only has solutions x=3. The denominators are zero at x=3, so these solutions are extraneous. If the right side is subtracted from the left side, the fractions can be combined to give ...
((x+3) -6)/((x -3)(x +3)) = 0 ⇒ 1/(x+3) = 0 . . . . no solution
