What is the product? (x2+3y+7) [yy+1)

PLEASE HURRY!!!!! What is the turning point of the graph of f(x) = |x – 4| ?

viva [34]

$f(x)=|x-4|\\
f'(x)=\text{sgn} (x-4)\\
\text{sgn} (x-4)=0\\
x-4=0\\
x=4\\\\

$

For $x$ $f'(x)$ .
For $x>4$ $f'(x)>0$ .
$\Downarrow\\
$
At $x=4$ there's turning point, which is equal $|4-4|=|0|=0$ .

8 0

3 years ago

Solve and plot your answer on the number line below: 7.08 + 2x=15.96

NemiM [27]

7.08 + 2x =15.96
-7.08. -7.08
-------------------------
2x=8.88 divide each by 2

X=4.44
You would plot it in between 4 and 5 but on one of the line marks depending on what there value are

6 0

3 years ago

Subtract \frac{3n+2}{n-4}-\frac{n-6}{n+4}

aalyn [17]

(3n+2)/(n-4) - (n-6)/(n+4)

common denominator (n-4)(n+4)

{(n+4)(3n+2)-(n-4)(n-6)}/{(n-4)(n+4)}

Use the foil method:

{(3n²+14n+8)-(n²-10n+24)}/{(n-4)(n+4)}

distribute negative sign:

{(3n²+14n+8-n²+10n-24)}/{(n-4)(n+4)}

subtract:

(2n²+24n-16)/{(n-4)(n+4)}

take out 2:

2{n²+12n-8}/{(n-4)(n+4)}

4 0

3 years ago

Read 2 more answers

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.