2 years ago

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Mathematics

3 answers:

Eva8 [605]2 years ago

4 0

Answer:

Thx for the points

Step-by-step explanation:

Vaselesa [24]2 years ago

3 0

OooOOooOoOo sweet i luv lofi

Foxo Boi2 years ago

0 0

E...............

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Larry is building a frame around a rectangular picture. He will need 42 inches of wood to do the entire frame. If the frame is 1

Anton [14]

The width would be 9 inches. Here's how I got the answer:

Double 12

Subtract from total

Take your difference and divide by two.

That will give you the width.

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2 years ago

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Svetach [21]

Answer:

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Step-by-step explanation:

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2 years ago

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f f(x)=2x f ( x ) = 2 x and g(x)=x2+3 g ( x ) = x 2 + 3 , find (g ∘ f)(x) ( g ∘ f ) ( x ) . Question 2 options: 2x2+3 2 x 2 + 3

Deffense [45]

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2 years ago

If A=(9,18) and B=(1,12), what is the length of AB?

Fynjy0 [20]

Use the distance formula?

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3 years ago

Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

EleoNora [17]

Start by multiplying each side of the inequality by $\sqrt{x + 1}$ and simplifying:

$2 \sqrt x + \frac{1}{\sqrt{x+1}} \leq 2 \sqrt{x + 1}$
$(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})$
$2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)$
$2 \sqrt{x^2 + x} + 1 \leq 2x + 2$
$2 \sqrt{x^2 + x} \leq 2x + 1$
$\sqrt{x^2 + x} \leq x + \frac{1}{2}$

From here, we can square both sides to get

$x^2 + x \leq (x + \frac{1}{2})^2$
$x^2 + x \leq x^2 + x + \frac{1}{4}$
$0 \leq \frac{1}{4}$ , which is always true.

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2 years ago