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3 years ago

2n > -20 Please help N=?

Mathematics

2 answers:

serg [7]3 years ago

6 0

I just need points but I hope you do good

Wewaii [24]3 years ago

3 0

Answer:

The value of x>-10

Step-by-step explanation:

The interval notation if you need it is (-10,♾)

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If y= 3x + 6, what is the minimum value of (x^3)(y)?

Rainbow [258]

Answer:

Given the statement: if y =3x+6.

Find the minimum value of $(x^3)(y)$

Let f(x) = $(x^3)(y)$

Substitute the value of y ;

$f(x)=(x^3)(3x+6)$

Distribute the terms;

$f(x)= 3x^4 + 6x^3$

The derivative value of f(x) with respect to x.

$\frac{df}{dx} =\frac{d}{dx}(3x^4+6x^3)$

Using $\frac{d}{dx}(x^n) = nx^{n-1}$

we have;

$\frac{df}{dx} =(12x^3+18x^2)$

Set $\frac{df}{dx} = 0$

then;

$(12x^3+18x^2) =0$

$6x^2(2x + 3) = 0$

By zero product property;

$6x^2=0$ and 2x + 3 = 0

⇒ x=0 and x = $-\frac{3}{2} = -1.5$

then;

at x = 0

f(0) = 0

and

x = -1.5

$f(-1.5) = 3(-1.5)^4 + 6(-1.5)^3 = 15.1875-20.25 = -5.0625$

Hence the minimum value of $(x^3)(y)$ is, -5.0625

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3 years ago

What number has factors of 2 and 3 and 12 and 18 as multiples?

Firlakuza [10]

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djyliett [7]

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Explain how bacteria are both helpful and harmful to out bodies

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Answer:

Step-by-step explanation:

Bacteria is helpful because it helps protect the cells in your intestines. Its, harmful because itI can cause a disease in your lungs.

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3 years ago

A square on a coordinate plane has the points A(0, 0), B(0, 5), C(5, 5) and D(5, 0). If Square ABCD is rotated about the origin

Georgia [21]

Answer:

B. 5 units

Step-by-step explanation:

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, dilation, rotation or translation.

If a point X(x,y) is rotated about the origin 180 degrees clockwise the new point is at X'(-x, -y).

If the square with vertices at A(0, 0), B(0, 5), C(5, 5) and D(5, 0) is rotated about the origin 180 degrees clockwise the new points are at A'(0, 0), B'(0, -5), C'(-5, -5), D'(-5, 0)

A square has four equal sides. The distance between two points $X(x_1,y_1)\ and\ Y(x_2,y_2)$ is:

$|XY|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Hence:

$|A'B'|=\sqrt{(0-0)^2+(-5-0)^2} =5$

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