<img src="https://tex.z-dn.net/?f=1%20%5Cdiv%201%20%2B%20a%20%20%7Bl%7D%20-%20%7Bm%7D%20%20%2B%201%20%5Cdiv%201%20%2B%20%20%7Bm%

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3 years ago

7D%20-%20%7Bl%7D%20" id="TexFormula1" title="1 \div 1 + a {l} - {m} + 1 \div 1 + {m} - {l} " alt="1 \div 1 + a {l} - {m} + 1 \div 1 + {m} - {l} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

ziro4ka [17]3 years ago

6 0

Answer:

Using BODMAS

first solve division, then go for addition or subtraction

$1 \div 1 + al - m + 1 \div 1 + m - l\\ =(1 \div 1) + al - m +( 1 \div 1) + m - l\\ =1 + al - m + 1 + m - l\\ = 2 + al -l\\=2 + l(a - 1)$

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The endpoints of directed line segment PQ have coordinates of P(-7, -5) and Q(5,3). What are the coordinates of point A, on PQ,t

Kamila [148]

Answer:

Therefore the co-ordinate of the point A is (-4,-3)

Step-by-step explanation:

If A(x₁,y₁) and B(x₂,y₂) is divided by a point in ration m:n internally.

Then the co-ordinate of the point is $(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_2}{m+n})$ .

Given two points are P(-7,-5) and Q(5,3).

A point A on PQ, that divides PQ into a ratio of 1:3.

Since the point A lies on the PQ.

Therefore it divides the line segment PQ internally.

Here x₁= -7,y₁= -5 , x₂= 5,y₂ =3, m=1,n=3

Therefore the co-ordinate of the point A is

$(\frac{5.1+(-7).3}{1+3},\frac{3.1+(-5).3}{1+3})$

$=(\frac{-16}{4},\frac{-12}{4})$

=(-4,-3)

6 0

3 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.