Which expression represents the number 13,809 written in expanded form?

The polynomial below is a perfect square trinomial of the form A^2-2AB+B^2. 9x^2-18x+36

baherus [9]

No, It's not.

coefficients for perfect square would be:
3² - 2(3)(6) + 6² = 9 - 36 + 36

8 0

3 years ago

Read 2 more answers

Area for rectangle in the coordinate plane. IXL Geometry help pls !

Sidana [21]

Answer:

Step-by-step explanation:

The co-ordinates of the rectangle are:

A (8, 6)

B (-4, -10)

C (-8, -7)

D (4, 9)

The Area of Rectangle is Given by: <em>length x breadth </em>= AB x BC

Lengths of AB and BC can be found by distance formula

<em>length</em>(AB) = $\sqrt{(x_{2} - x_{1} )^{2} + (y_{2} - y_{1} )^{2}}$ = $\sqrt{(-4-8)^{2} + (-10-6)^{2}} = 20$

Similarly,

<em>length</em>(BC) = 5

Area of Rectangle = AB x BC = 20 x 5 = 100 square units

5 0

3 years ago

If two sides of a triangle are 5 and 7 inches longer than the third side and the perimeter measures 21, find the length of each

aleksandr82 [10.1K]

Hello

x
x +5
x + 7

x + x + 5 + x + 7 = 21
3 x = 21 - 12
3 x = 9
x = 3

3 , 5 et 7 pouces

7 0

3 years ago

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

3 years ago

If f(x)=4^x, find f(4)

lina2011 [118]

Answer:

$\huge\boxed{f(4) = 256}$

Step-by-step explanation:

<u>Given function is:</u>

$f(x) = 4^x$

Put x = 4

So,

$f(4) = 4^4\\\\f(4) = 256\\\\\rule[225]{225}{2}$

Hope this helped!

6 0

2 years ago