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vampirchik [111]
3 years ago
14

Find the factorization of the polynomial below . 64x^2 +48x+9

Mathematics
2 answers:
alukav5142 [94]3 years ago
7 0

Answer:

(8x+3)²

Step-by-step explanation:

The polynomial has perfect squares on wither end of the expression including 64 and 9. This means it is likely factoring to (8x +3)². TO check this we will multiply it out.

(8x+3)(8x+3) = 64x² + 24x + 24x + 9 = 64x² + 48x + 9

valina [46]3 years ago
3 0

64x^2 +48x +9

MODE ×3 = EQN = DEGREE 2

(8x+3)(8x+3)

x= -3/8

answer = (8x+3)²

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Given the circumference of a circle is 8π centimeters, calculate the area of the circle in terms of π.
arsen [322]

Answer:

16π cm²

Step-by-step explanation:

circumference = 2πr

Area = πr²

We need to determine r to calculate the area

2πr = 8π

Divide both sides of the equation by 2π to find r

r = 4

Now, we determine area.

area =  πr² =   π4² = 16πcm²

6 0
3 years ago
Expand.If necessary, combine like terms (2x-3)(2x-3)
ivanzaharov [21]

Answer:

4x^2-12x+9 hope this helps

Step-by-step explanation:

4 0
3 years ago
Find the absolute maximum and minimum values of f on the set D. f(x,y)=2x^3+y^4, D={(x,y) | x^2+y^2<=1}.
castortr0y [4]

Answer:

absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.

Step-by-step explanation:

We compute,

$ f_x = 6x^2, f_y=4y^3 $

Hence, $ f_x = f_y = 0 $  if and only if (x,y) = (0,0)

This is unique critical point of D. The boundary equation is given by

$ x^2+y^2=1$

Hence, the top half of the boundary is,

$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}

On T we have, $ f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1  \text{ for}\ -1 \leq x \leq 1$

We compute

$ \frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$

0 if and  only if x=0, x= 1/2 or x = -2.

We disregard  $ x = -2 \notin [-1,1]$

Hence, the critical points on T are (0,1) and $(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$

On the bottom half, B, we have

$ f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$

Therefore, the critical points on B are (0,-1) and $( 1/2, -\sqrt3/2)  

It remains to  evaluate f(x, y) at the points $ (0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\  (\pm1, 0)$ .

We should consider  latter two points, $(\pm1,0)$, since they are the boundary points for the T and also  B. We compute $ f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2 $

We conclude that the  absolute maximum = f(1, 0) = 2

And the absolute minimum = f(−1, 0) = −2.

6 0
3 years ago
Help I'm practicing ​
serious [3.7K]
It’s definitely either A or C
8 0
3 years ago
Read 2 more answers
Which expression is equivalent to 5/6c + 76c, if c+ 0?
mestny [16]

Answer:

0

Step-by-step explanation:

If c=0

that expression also equals 0

6 0
3 years ago
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