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andreyandreev [35.5K]
2 years ago
13

Factorise:p/4-q2/16 pls help me with it

Mathematics
2 answers:
Annette [7]2 years ago
7 0

Answer:

1/4 ( p - 1/4 q^2).

Step-by-step explanation:

p/4 - q2/16

The greatest common factor is 1/4 so we have the answer:

1/4 ( p - 1/4 q^2).

Varvara68 [4.7K]2 years ago
3 0

Answer:

\frac{1}{4} (p-\frac{2q}{4} )

Step-by-step explanation:


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A) How many ways can 2 integers from 1,2,...,100 be selected
Anna007 [38]

Answer with explanation:

→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

7 0
3 years ago
Keith thinks of a number. When he multiples the number by 6 and subtracts 19.85 from the product, he gets 29.77. Find the number
kirill115 [55]

let the number be X

6x-19.85=29.77

6x=29.77+19.85

6x=49.62

x=8.27

Hope that helps :)

-Asmaa Ghazzawi

8 0
3 years ago
Mister Rogers is fencing another new rectangular garden in his neighborhood. One side of the garden faces the road and needs to
Oksanka [162]

Answer:

length of the pretty side  and length of the side oppositte to the pretty side  =   37.91 ft

length of the other two sides  = 27.52 ft

Step-by-step explanation:

The mathematical problem is:

Max A = b1*h

subject to: 35*b1 + 18*(2*h + b2) <= 3000

Where

A: area of the garden

b1: length of the pretty side

b2: length of the side oppositte to the pretty side

h: length of the other two sides

Replacing with b1 = b2 and taking only the equality sign in the restriction (in the maximum all the money will be spent), we get:

35*b1 + 18*(2*h + b1) = 3000

35*b1 + 36*h + 18*b1 = 3000

53*b1 + 36*h = 3000

b1 = 3000/53 - (36/53)*h

Substituing in Area's formula  

A = (3000/53 - (36/53)*h)*h

A = (3000/53)*h - (36/53)*h^2

In the maximum, the derivative of A is equal to zero

dA/dh = 3000/53 - 2*(36/53)*h =

3000/53 - 72/35*h = 0

h = (3000/53)*(35/72)

h = 27.52 ft

then,

b1 = 3000/53 - (36/53)*27.52

b1 = 37.91 ft =b2

3 0
3 years ago
Planes X and Y intersect at a right angle. Line A B and Line C G lie in plane X and do not intersect. Line R S lies in plane Y.
RSB [31]

Answer:

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Step-by-step explanation:

6 0
2 years ago
Y = 10.5 x + 5<br> Please help me solve this equation.<br> Solve for x.
Schach [20]
What is the context of the question
8 0
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