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3 years ago

PLZ HELP 70 POINTS!!!!!

Mathematics

2 answers:

marin [14]3 years ago

8 0

5^(x+7)=(1/625)^(2x-13)

We move all terms to the left:

5^(x+7)-((1/625)^(2x-13))=0

Domain of the equation: 625)^(2x-13))!=0

x∈R

We add all the numbers together, and all the variables

5^(x+7)-((+1/625)^(2x-13))=0

We multiply all the terms by the denominator

(5^(x+7))*625)^(2x+1-13))-((=0

We add all the numbers together, and all the variables

(5^(x+7))*625)^(2x-12))-((=0

We add all the numbers together, and all the variables

(5^(x+7))*625)^(2x=0

not sure if this is right :/

QveST [7]3 years ago

8 0

Hey there! the answer to this equation is

x=1/3125

Explanation

You move all the terms to the left:

5^x-(1/625)^(2x-13)=0

Then add all the numbers altogether, and the variables too

5^x-(+1/625)^(2x-13)=0

Now get rid of the parentheses

5^x-1/625^2x-13=0

Multiply all of them by the denominator

5^x*625-1 .2x-13=0

Finally , move all terms to the left, and all other terms to the right

This should be your final answer

3125x=1

Hope this helped you and if im wrong, correct me.

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Please help on number 3

kobusy [5.1K]

Answer:

Its the first option.

Step-by-step explanation:

3x + y = 5

5x - 4y = -3

From equation 1:

y = - 3x + 5

Substituting for y:

5x - 4(-3x + 5) = -3

5 0

2 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.