hectictar

area of square  =  (side length)²

                                                          Plug in  325 m²  for  area and solve for side length.

325 m²  =  (side length)²

                                                          Take the positive square root of both sides.

√[325 m²]  =  side length

5√13 m  =  side length

perimeter of square  =  4 * side length

                                                                 Plug in  5√13 m  for  side length.

perimeter of square  =  4 * 5√13 m

perimeter of square  =  20√13  m

Hey! Welcome to the forum! 

Maybe it is supposed to be

f(x)  =  2(2)^x - 4        ???

If so...

f(0)   =   2(2)⁰ - 4   =   2(1) - 4   =   2 - 4   =   -2     So the graph passes through  (0, -2) .

f(1)   =   2(2)¹ - 4   =   2(2) - 4   =   4 - 4   =   0     So the graph passes through  (1, 0) .

The first graph passes through  (0, -2)  and  (1, 0)  , so it looks right.

-3x⁴ + 27x² + 1200  =  0

First, let's divide each term by  -3  .

x⁴ - 9x² - 400  =  0

Now, notice that we can write  x⁴  as  (x²)² .

(x²)² - 9x² - 400  =  0

And this factors as...

(x² - 25)(x² + 16)  =  0            Set each factor equal to zero.

x² - 25  =  0          or          x² + 16  =  0

x²  =  25               or           x²  =  -16

Now we can solve each equation for  x  by taking the ± square root of both sides.

x  =  ± 5          or          x  =  ± 4i

So all the zeros are   5 ,  -5 ,  4i , and  -4i

f(x)  =  ax^r

f(2)  =  1

a(2)^r  =  1                  Divide both sides by  2^r .

a  =  1 / ( 2^r)

f(32)  =  4

a(32)^r  =  4              Divide both sides by  32^r .

a  =  4 / ( 32^r)

Set these two values of  a  equal to each other.

1 / ( 2^r)  =  4 / (32^r)         Cross multiply.

32^r  =  4 * 2^r                   And we can write  32  and  4  as powers of  2 .

(2^5)^r  =  2^2 * 2^r

2^(5r)  =  2^(2 + r)

5r  =  2 + r

4r  =  2

r  =  1/2

the side length of the cube   =   \(\sqrt[3]{216}\)  cm   =   6 cm

So we know...

volume of water + volume of cube   =   10 * 12 * 6  cm³   =   720 cm³

volume of water +  216 cm³   =   720 cm³

volume of water   =   504  cm³

10 cm * 12 cm * original height   =   504  cm³

original height   =   504 / 120  cm   =   4.2 cm           

These angles are supplementary, so...

(x + 1)°  +  (9x - 21)°   =   180°

x + 1 + 9x - 21   =   180

                                         Combine  x  and  9x  to get  10x , and combine  1  and  -21  to get  -20 .

10x - 20  =  180

                                         Add  20  to both sides of the equation.

10x   =   200

                                         Divide both sides of the equation by  10 .

x   =   20

So....

x + 1   =   20 + 1   =   21

9x - 21   =  9(20) - 21   =   159

area of triangle  =  (1/2) * base * height

The problem tells us...

base  =  27 mm

height  =  48 mm

So to find the area, just plug in the values we know for base and height.

area of triangle  =  (1/2) * 27 * 48      mm²        ....Can you finish it?  

area of trapezpoid   =   (first base + second base) / 2  *  height

area of trapezpoid   =   (6 + 9) / 2  *  4

area of trapezpoid   =   15 / 2  *  4

area of trapezpoid   =   60 / 2

area of trapezpoid   =   30       cm²

a)  A line through the origin, which is  (0, 0) , and  (10, 11)  has a slope of

[11 - 0] / [10 - 0]   =   11/10

and a line parallel to this also has a slope of  11/10 .

b)  A line that passes through  (3, 8)  and  (-1, 9)  has a slope of

[9 - 8] / [-1 - 3]   =   1 / -4   =   -1/4

and the slope of a perpendicular line  =  the negative reciprocal of  -1/4  =  4

c)   A line parallel to the  x-axis  will not "rise" at all between any two points.

its slope  =  rise / run  =  0 / run  =  0