Name the lengths of the side of three rectangles that have perimeters of 14 units

HELP ASAP!!! PLEASE CHECK ALL OF THEM!

pochemuha

Answer:

2.

A. (P+h)(x)

2x/x+4 (x-1) + x/x-1 (x+4)

2x^2-1/x^2-4

+

X^2+4/x^2-4

= 3x^2+3/x^2-4

B. (F-g)(x)

X^2-7x+6-x - 6

= x^2 -8x

C. (Fg)(x)

(X^2-7x+6)(x-6)

= x^3-13x^2+48x-36

D. (H/p)(x)

X/x-1 / 2x/x+4

X/x-1 / x+4/2x

= X^2+4x/2x^2-2x

3.

A. (F+g)(3)

X^2+1 + x-4

3^2+1 + 3-4

10 -1

= 9

B. (f-g)(0)

X^2+1 - x-4

0+1 -0-4

1-4

= -3

C. (Fg)(-k)

(X^2+1) (x-4)

(-k^2+1) (-k-4)

K^3+4k^2-k-4

D. (F/g)(k-2)

X^2+1 /x-4

K-2^2+1 / k-2 -2

= K^2-4k+5 / k-4

Step-by-step explanation:

8 0

3 years ago

$190,258.50; $152,698.00; $122,753.00; $220,523.00; $231,951.00. What is the average of these five amounts?

Yakvenalex [24]

The average of those five amounts is 183,636.7
AVeraging is adding all of the numbers together with the dividing by a number of numbers you have.

6 0

3 years ago

I have 93, 80 and 94 my homework is worth 10% so what us my grade?

Minchanka [31]

I think about 26.7
or something around there

5 0

3 years ago

Read 2 more answers

How do I solve this?

Len [333]

5 Girls + 3 Boys =8 (sample size)

Probability of choosing 1 boy P(1 B) = 3/8

Probability of NOT CHOOSING ANY BOY = 1-3/8 =5/8

Now apply the binomial probability to choose 3 Boys:

8C3(3/8)³(5/8)⁵ ==>P(All boys) = 0.28

7 0

3 years ago

I need help please, I am so confused.

vazorg [7]

Answer:

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

Step-by-step explanation:

x-intercepts are when the curve intercepts the x-axis, so when y =0.

Therefore, to find the x-intercepts, substitute y = 0 and solve for x.

The vertex is the turning point: the minimum point of a parabola that opens upward, and the maximum point of the parabola that opens downward. As a parabola is symmetrical, the x-coordinate of the vertex is the midpoint of the x-intercepts.

Equation: $y=x(x-2)$

$\implies x(x-2)=0$

$\implies x=0$

$\implies (x-2)=0 \implies x=2$

Therefore, the x-intercepts are x = 0 and x = 2

The midpoint of the x-intercepts is x = 1, so the x-coordinate of the vertex is x = 1

Equation: $y=(x-4)(x+5)$

$\implies (x-4)(x+5)=0$

$\implies (x-4)=0 \implies x=4$

$\implies (x+5)=0 \implies x=-5$

Therefore, the x-intercepts are x = -5 and x = 4

The midpoint of the x-intercepts is x = -0.5, so the x-coordinate of the vertex is x = -0.5

Equation: $y=-5x(3-x)$

$\implies -5x(3-x)=0$

$\implies -5x=0 \implies x=0$

$\implies (3-x)=0 \implies x=3$

Therefore, the x-intercepts are x = 0 and x = 3

The midpoint of the x-intercepts is x = 1.5, so the x-coordinate of the vertex is x = 1.5

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

4 0

2 years ago