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2 years ago

6x – 4 < –34 or 3x + 10 > 10

2 answers:

5 0

1. 6x-4< -34, or 3x+10 > 10.

6x-4 < -34. 
6x < -34 + 4 
6x < -30 
X < -5. 

3x + 10 > 10. 
3x > 10 - 10 
3x > 0 
X > 0. 

2. -2 <= 2x-4 < 4. 
Add 4 to each end and the center: 
-2 + 4<= 2x-4+4 < 4+4. 
2 <= 2x < 8 
Divide each end and center by 2: 
1 <= X < 4. 

The compound inequality is actually 2 
inequalities and can be solved separately: 

2x-4 => -2, and 2x-4 < 4. 

2x-4 => -2. 
2x => -2+4 
2x=> 2 
X => 1. 

2x-4 < 4. 
2x < 4+4 
2x < 8 
X < 4. 

Solution: 1< = X < 4. or X => 1, and X<4 

 TO CHECK: 

1. Substitute any number < -5 for X. 
Eq2: Subst. any number > 0 for X.

kvv77 [185]2 years ago

5 0

Left side: x < -5

Right side: x > 0

Not sure if you want to graph it on a number line or a coordinate plane

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Please answer the geometry question

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300 feet

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2 years ago

Translate the following written expression into an algebraic expression:

hammer [34]

Answer:

(y-5) ÷ 2

Step-by-step explanation:

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2 years ago

What is the equation of the oblique asymptote? h(x) = x² – 3x - 4/x + 2

NNADVOKAT [17]

Simplifying h(x) gives

h(x) = (x² - 3x - 4) / (x + 2)

h(x) = ((x² + 4x + 4) - 4x - 4 - 3x - 4) / (x + 2)

h(x) = ((x + 2)² - 7x - 8) / (x + 2)

h(x) = ((x + 2)² - 7 (x + 2) - 14 - 8) / (x + 2)

h(x) = ((x + 2)² - 7 (x + 2) - 22) / (x + 2)

h(x) = (x + 2) - 7 - 22/(x + 2)

h(x) = x - 5 - 22/(x + 2)

An oblique asymptote of h(x) is a linear function p(x) = ax + b such that

$\displaystyle \lim_{x\to\pm\infty} h(x) - p(x) = 0$

In the simplified form of h(x), taking the limit as x gets arbitrarily large, we obviously have -22/(x + 2) converging to 0, while x - 5 approaches either +∞ or -∞. If we let p(x) = x - 5, however, we do have h(x) - p(x) approaching 0. So the oblique asymptote is the line y = x - 5.

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2 years ago

AACB = ADCE. If AC = 5 and BC = 7 CD = [ ? ). PLS HELP LOL PLS

FromTheMoon [43]

Answer:

CD = 5

Step-by-step explanation:

AC = 5

BC = 7

∆ACB ≅ ∆DCE, therefore,

AC = CD,

BC = CE, and,

AB = DE

Thus,

AC = CD = 5

CD = 5

6 0

2 years ago

Using triangle A B C as the pre-image and the origin as the center of dilation, which coordinate point is not a coordinate of a

Ostrovityanka [42]

Answer:

The correct option is:

begin ordered pair 4 comma 12 end ordered pair i.e. (4,12).

Step-by-step explanation:

We are given the coordinates of the point A,B and C of the triangle ΔABC as:

A(-1,3) , B(2,1) and C(-2,-1).

Now it is given that there is dilation of this triangle about the origin with a scale factor of 4.

This means that the change in the coordinates after the transformation is given by the rule:

(x,y) → (4x,4y)

As A → A' , B → B' and C → C'

Hence the coordinates A',B' and C' are given by:

A' are (4×-1,4×3)=(-4,12)

B' are (4×2,4×1)=(8,4)

and C' are (4×-2,4×-1)=(-8,-4).

Hence the coordinate point which is not a coordinate of dilation is:

begin ordered pair 4 comma 12 end ordered pair i.e. (4,12)

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3 years ago