Write 4/5 as an equivalent fraction in 3 different forms.

On your paper, graph these coordinates:<br> (0, 1), (2, 5), (3,7)

KATRIN_1 [288]

The equation of the line is y = 2x + 1

4 0

3 years ago

The width of a rectangle is the sum of the length and 5. The area of the rectangle is 36 units. What id the width, in units, of

Alexxandr [17]

Answer:

Step-by-step explanation:

Let length = x

Width = x + 5

Area = 36

Area = length × width

36 = x(x + 5)

36 = x^2 + 5x

x^2 + 5x - 36 = 0

x^2 + 9x - 4x -36 =0

(x^2 + 9x) - (4x + 36) = 0

x(x + 9) - 4(x + 9) = 0

(x -4)(x+9) = 0

x - 4 = 0 or x + 9 = 0

x = 4 or -9

But x can only be positive because it is a length

x = 4unit = length

Width = x + 5 = 4 +5 = 9unit

7 0

3 years ago

Find the inverse of P(x)=-15x^2+350x-2000 and state it’s domain.

natima [27]

Answer:

The function given is:

P(x) = 15x² + 350x - 2000

or

y = 15x² + 350x - 2000

Interchange x and y:

x = 15y² + 350y - 2000

x = (y√15)² + 2(y√15)(45.19) + (45.19)² - (45.19)² - 2000

x = ( y√15 + 45.19)² - 4042.14

( y√15 + 45.19)² = x + 4042.14

Take square root:

y√15 + 45.19 = (√(x + 4042.14))

y√15 = (√(x + 4042.14)) - 45.19

$y = \sqrt{\frac{x+4042.14}{15}}-11.67$

$f^{-1}(x) = \sqrt{\frac{x+4042.14}{15}}-11.67$

The minimum value of x, for which function is defined and real, is x= -4042.14

Hence, the range of inverse is x ≥ -4042.14

4 0

3 years ago

In the diagram below, R is located at (24,0), N is located at (12,18), T is located at (12,6), and E is located at (18,15). Assu

julsineya [31]

The midpoint of a segment divides the segment into equal halves

The coordinates of K are: $\mathbf{K = (12,12)}$
The coordinates of I are: $\mathbf{I = (24,24)}$
The coordinates of B are: $\mathbf{B = ( 30,33 )}$

The given parameters are:

$\mathbf{R =(24,0)}$

$\mathbf{N =(12,18)}$

$\mathbf{T =(12,6)}$

$\mathbf{E =(18,15)}$

K is the midpoint of N and T.

So, we have:

$\mathbf{K = (\frac{N_x + T_x}{2},\frac{N_y + T_y}{2})}$

This gives

$\mathbf{K = (\frac{12 + 12}{2},\frac{18+ 6}{2})}$

$\mathbf{K = (12,12)}$

E is the midpoint of T and I.

So, we have:

$\mathbf{E = (\frac{I_x + T_x}{2},\frac{I_y + T_y}{2})}$

This gives

$\mathbf{(18,15) = (\frac{I_x + 12}{2},\frac{I_y+ 6}{2})}$

Multiply through by 2

$\mathbf{(36,30) = (I_x + 12,I_y+ 6)}$

By comparison

$\mathbf{I_x + 12 = 36.\ I_y + 6 =30}$

So, we have:

$\mathbf{I_x= 24.\ I_y =24}$

Hence, the coordinates of I are:

$\mathbf{I = (24,24)}$

I is the midpoint of E and B.

So, we have:

$\mathbf{I = (\frac{E_x + B_x}{2},\frac{E_y + B_y}{2})}$

This gives

$\mathbf{(24,24) = (\frac{18 + B_x}{2},\frac{15 + B_y}{2})}$

Multiply through by 2

$\mathbf{(48,48) = (18 + B_x,15 + B_y)}$

By comparison

$\mathbf{18 + B_x = 48,\ 15 + B_y = 48 }$

So, we have:

$\mathbf{B_x = 30,\ B_y = 33 }$

Hence, the coordinates of B are:

$\mathbf{B = ( 30,33 )}$