Algebra help ASAP :))))

Mathematics

1 answer:

never [62]3 years ago

3 0

G(x) = (1/2x)²

Place points at:
(-2, 2)
(-1, 0.5)
(0, 0)
(1, 0.5)
(2, 2)

The parabola should look something like this.

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You are planning to put a new digital flat t.v on a wall that is 12ft long and 9ft high. the digital t.v has a diagonal of 72 in

Pie

Area of wall = 9 * 12 = 108 ft^2
Width of TV = x
Lenght = 2x
Diagonal of TV = 72 " = 6 ft

x^2+(2x)^2=6^2
x^2+4x^2=36
5x^2=35
x^2=36/5
x=sqrt(36/5)
Width = 2.68 ft.
Length = 2.68 * 5.36 = 14.36 ft^2
Area of wall - area of tv = 108 - 14.36 = 93.64 ft

7 0

2 years ago

Completely factorize (2x+3)^-2 + 2x(2x+3)^-3

Alexxx [7]

Hello,

$\dfrac{1}{(2x+3)^2} + \dfrac{2x}{(2x+3)^3} \\

= \dfrac{(2x+3)+2x}{(2x+3)^3}\\

= \dfrac{4x+3}{(2x+3)^3}$

4 0

2 years ago

A linear function that represents the number of video games purchased at an electronics store is compared to a different linear

serg [7]

Answer:

See there y-intersept and if slope is negative or positive to see if they intersect

7 0

2 years ago

Can I please get some help on this...I have tried to many times...

Bezzdna [24]

Answer: A

Step-by-step explanation:

First, the problem is g(f(x)). You would plug in f(x) wherever you see an x in g(x). To find the domain, you take the bottom function, and set it equal to 0.

$\sqrt{x-2} =0$

When you solve that, you get x=2. You know your domain is x≥2, but there is as asymptote at x=11. That means the graph never reaches x=11, but gets very close. You find that by setting the entire equation equal to 0 and solve from there.

5 0

3 years ago

And slope of the linear equation y+8=3(x−5)<br> ?

Triss [41]

Answer:

slope: 3; vertex: (5, -8)

Step-by-step explanation:

Compare the given y+8=3(x−5) to the slope-vertex form y - k = m(x - h). We see that the slope, m, must be 3 and the vertex must be (h, k): (5, -8).

7 0

2 years ago