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2 years ago

Solve the inequality:12+x≥3(x-6)

Mathematics

1 answer:

SVETLANKA909090 [29]2 years ago

8 0

Hello there! :)

The first step is to use the Distributive Property.

It states that

a(b+c)=ab+ac

Let's distribute 3, using this property:

3(x-6)=3x-18

Now, our inequality looks like so:

12+x≤3x-18

The next step is to move the variables to the left, using the opposite operation:

12+x-3x≤-18

12-2x≤-18

Now, move all the numbers to the right, using the opposite operation:

-2x≤-18-12

-2x≤-30

Divide both sides by -2 to isolate x.

Remember that when we divide by a negative variable, we change the inequality sign:

x≥15

Therefore, x≥15

Hope it helps.

~A lonely teen who helps others with a smile

-Grace :-)

Good luck.

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2 years ago

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bija089 [108]

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3 years ago

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3 years ago

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3 years ago

Solve y=f(x) for x. Then find the input when the output is -3.

DochEvi [55]

Answer:

Please check the explanation

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:\left(x-5\right)^3-1$

Given that the output = -3

i.e. y = -3

now substituting the value y=-3 and solve for x to determine the input 'x'

$\:\:y=\:\left(x-5\right)^3-1$

$-3\:=\:\left(x-5\right)^3-1\:\:\:$

switch sides

$\left(x-5\right)^3-1=-3$

Add 1 to both sides

$\left(x-5\right)^3-1+1=-3+1$

$\left(x-5\right)^3=-2$

$\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

Thus, the input values are:

$x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}$

And the real input is:

$x=-\sqrt[3]{2}+5$

$x=3.74$

4 0

3 years ago

Solve the inequality:12+x≥3(x-6)​

Solve the inequality:12+x≥3(x-6)