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3 years ago

-12-x>-12+4x put the steps too pls thanks

Mathematics

1 answer:

andrey2020 [161]3 years ago

3 0

Answer:

x < 0

Step-by-step explanation:

Firstly put the values with variable x to one side,

; -12 - x > -12 + 4x

; -12 + 12 > 4x + x

; 0 > 5x

then divide both sides by 5,

; (0 ÷ 5) > (5x ÷ 5)

; 0 > x

; Hence the solution to the inequality above,

; x < 0...always remember to reverse the sign when you change the side to which you put the variable of x,

Therefore...( x < 0 )

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Someone just answer this math question thingie lol. you’ll get 33 points G

Stels [109]

1) $(-2x+4)+(x-6)$ is simplified into $=-x-2$

2) $(uv-3)+(4uv+1)$ is simplified into $5uv-2\\$

3) $(11x-8)-(2x+3)$ is simplified into $-9x-11$

4) $(x^2-4)-(x-2)$ is simplified into $x^2-x-2$

5) $(12-3t-7t^2)+(1+3t-t^2)$ is simplified into $-8t^2+13$

Step-by-step explanation:

We need to solve the polynomials.

1) $(-2x+4)+(x-6)$

Solving:

$(-2x+4)+(x-6)\\=-2x+4+x-6\\Combining\,\,like\,\,terms:\\=-2x+x+4-6\\=-x-2$

So, $(-2x+4)+(x-6)$ is simplified into $=-x-2$

2) $(uv-3)+(4uv+1)$

Solving:

$(uv-3)+(4uv+1)$

Expanding:

$=uv-3+4uv+1\\Combining\,\,like\,\,terms:\\=uv+4uv-3+1\\=5uv-2\\$

So, $(uv-3)+(4uv+1)$ is simplified into $5uv-2\\$

3) $(11x-8)-(2x+3)$

Solving:

$(11x-8)-(2x+3)$

Expanding:

$=11x-8-2x-3\\Combining\,\,like\,\,terms:\,\,\\=11x-2x-8-3\\=-9x-11$

So, $(11x-8)-(2x+3)$ is simplified into $-9x-11$

4) $(x^2-4)-(x-2)$

Solving:

$(x^2-4)-(x-2)$

Expanding:

$=x^2-4-x+2\\Combining\,\,like\,\,terms:\\=x^2-x-4+2\\=x^2-x-2$

SO, $(x^2-4)-(x-2)$ is simplified into $x^2-x-2$

5) $(12-3t-7t^2)+(1+3t-t^2)$

Solving:

$(12-3t-7t^2)+(1+3t-t^2)$

Expanding:

$=12-3t-7t^2+1+3t-t^2\\Combining\,\,like\,\,terms:\\=12+1-3t+3t-7t^2-t^2\\=13+0t-8t^2\\=-8t^2+13$

So, $(12-3t-7t^2)+(1+3t-t^2)$ is simplified into $-8t^2+13$

Keywords: Solving Polynomials

Learn more about solving Polynomials at:

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Which graph represents f(x)=log3x 1?

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The correct option A: graph A, shows the logarithmic function;

f(x) = log₃x + 1

<h3>Explain the term logarithmic function?</h3>

The opposite of such an exponential function is a logarithmic function.
A log function and then an exponential function both use the same base.
An exponent is a logarithm. f(x) = bx is how the exponential function be expressed.
The formula again for logarithmic function is f(x) = log base b of x.

f(x) = log₃x + 1

When x = 1

f(1) = log₃1 + 1

f(1) = 0 + 1

f(1) = 1

(1, 1)

When x = 3

f(3) = log₃3 + 1

f(3) = 1 + 1

f(3) = 2

(3, 2)

Thus, the graph must contain the points (1, 1), (3, 2).

Therefore, graph A define the given logarithmic function;

f(x) = log₃x + 1.

To know more about the logarithmic function, here

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