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3 years ago

Favorite vine? giving brainliest <3

Mathematics

2 answers:

Naya [18.7K]3 years ago

5 0

Answer:lOoK aT aLl ThEsE cHiCkEnS

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svlad2 [7]3 years ago

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Answer:

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Solve for x in the equation 2 x squared + 3 x minus 7 = x squared + 5 x + 39.

Nastasia [14]

Answer:

$\implies x=1\pm \sqrt{47}$

Step-by-step explanation:

We want to solve for x in $2x^2+3x-7=x^2+5x+39$

You need to group and combine like terms and write in standard form:

$2x^2-x^2+3x-5x-7-39=0$

$\imples x^2-2x-46=0$

By comparing to $ax^2+bx+c=0$ , we a=1,b=-2 and c=-46

The solution can be obtained using the quadratic formula.

$x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

We substitute the coefficients to get:

$x=\frac{--2\pm \sqrt{(-2)^2-4*1*-46} }{2*1}$

$x=\frac{2\pm \sqrt{4+4*46} }{2}$

$x=\frac{2\pm \sqrt{4*47} }{2}$

$x=\frac{2\pm2\sqrt{47} }{2}$

$\implies x=1\pm \sqrt{47}$

The last choice is correct

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3 years ago

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5. <br> 56°<br> 94°<br> 4.5%<br> 1.5x

dedylja [7]

Answer:

Step-by-step explanation:

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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.