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3 years ago

Help ASAP!!!!!!!!!!!!

Mathematics

1 answer:

zepelin [54]3 years ago

7 0

Answer:

x is greater than or equal to -4

Step-by-step explanation:

I am not sure if I have to graph as well so I will just solve the inequality.

3x-12 greater than or equal to 7x+4

+12 +12

3x is greater than or equal to 7x+16

-7x -7x

-4x is greater than or equal to 16

-4x/-4 16/-4

x is greater than or equal to -4

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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:

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

$x=3.74$

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A square room has a tiled floor with 81 square tiles. How many tiles are along an edge of the room?

Sladkaya [172]

Answer:

36 tiles

Step-by-step explanation:

Because the room is square, the numbers of tiles along the edge of the room were multiplied by the same number, meaning that:

81 = x²

So to find x, we must square root both sides of the equation:

$\sqrt{81} = \sqrt{x^2}$

x = 9

So the length of one side is 9 tiles. Now to calculate the circumference, as each side of a square is the same length, we can multiply this number by 4:

9 * 4 = 36

So there are 36 tiles along the edge of the room.

Hope this helps!

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