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3 years ago

X<= -2 on a number line

Mathematics

1 answer:

ruslelena [56]3 years ago

7 0

Answer:

$x\le \:-2\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-2\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-2]\end{bmatrix}$

Please check the attached number line graph.

The number line clearly indicates that the graph is heading towards negative infinity from -2.

Step-by-step explanation:

Given the inequality expression

x ≤ -2

The inequality symbol ' ≤ ' means 'less than or equal to'.

Thus,

x ≤ -2 means x is less than or equal to -2.

In other words,

$x\le \:-2\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-2\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-2]\end{bmatrix}$

Please check the attached number line graph.

The number line clearly indicates that the graph is heading towards negative infinity from -2.

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Solve the system of equations

nikitadnepr [17]

The solution is A

x=2 y=-1

the solution is where the two lines intersect

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3 years ago

Read 2 more answers

What is the volume of a cylinder, in cubic centimeters, with a height of 8 centimeters

Arisa [49]

Answer:

1608.5 cm³

Step-by-step explanation:

Use the cylinder volume formula, V = $\pi$ r²h

If the diameter is 16 cm, then the radius is 8 cm.

Plug in the radius and height into the formula, and solve:

V = $\pi$ r²h

V = $\pi$ (8)²(8)

V = $\pi$ (64)(8)

V = 1608.5

So, to the nearest tenth, the volume of the cylinder is 1608.5 cm³

6 0

3 years ago

WHAT MONOMIAL MULTIPLIED BY -3X^3Y EQUALS 18X^9Y^4

Morgarella [4.7K]

-6x^3y^4
I think that is the answer but i don't know

5 0

3 years ago

Please solve this question

Alchen [17]

Answer:

(3, 4 )

Step-by-step explanation:

3x - y = 5 → (1)

x + y = 7 → (2)

adding the 2 equations term by term will eliminate y

4x + 0 = 12

4x = 12 ( divide both sides by 4 )

x = 3

substitute x = 3 into either of the 2 equations and solve for y

substituting into (2)

3 + y = 7 ( subtract 3 from both sides )

y = 4

solution is (3, 4 )

4 0

2 years ago

how do you rewrite : y=x^2-5x+9 in standard form using complying the square? I keep getting to : x^2-5x-25/4 but I’m not allowed

yawa3891 [41]

I presume with this you meant to say you want this equation in general form, also known as vertex form, which is y = a(x - h)² + k, instead of standard form, which is y = ax² + bx + c, which is what the original equation is already in.

So with completing the square, you first have to isolate the x-terms. To do this, subtract 9 on both sides of the equation:

$y-9=x^2-5x$

Next, we want to make the right side of the equation a perfect square. To find the constant of this soon-to-be perfect square, divide the x-coefficient by 2 and square that quotient. In this case:

$-5\div 2 = -\frac{5}{2}\\\\(-\frac{5}{2})^2=-\frac{5}{2}\times-\frac{5}{2}=\frac{25}{4}$

Now add 25/4 on both sides of the equation:

$y-9+\frac{25}{4}=x^2-5x+\frac{25}{4}$

Now to combine -9/1 and 25/4, they must have the same denominator, and to find it you must find the LCD, or lowest common denominator, of 1 and 4. To find it, list the multiples of both and the lowest one they share is their LCD. In this case, the LCD is 4. Multiply both sides of -9/1 by 4/4 and then add the numerators up:

$-\frac{9}{1}\times \frac{4}{4}=-\frac{36}{4}\\\\-\frac{36}{4}+\frac{25}{4}=-\frac{11}{4}\\\\y-\frac{11}{4}=x^2-5x+\frac{25}{4}$

Next, we need to factor the right side of the equation. Using this format of $a^2-2ab+b^2=(a-b)^2$ , we can factor it as:

$y-\frac{11}{4}=(x-\frac{5}{2})^2$

Now lastly, add both sides by 11/4, and your final answer will be $y=(x-\frac{5}{2})^2+\frac{11}{4}$ 

8 0

4 years ago