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2 years ago

PLEASE HELP ASAP ITS GEOMETRY

Mathematics

1 answer:

Gwar [14]2 years ago

4 0

The missing options that makes up the distance formula between two coordinates are; (x₂ - x₁)² and d²

How to find the distance between two coordinates?

The formula for distance between two coordinates is;

d = √[(y₂ - y₁)² + (x₂ - x₁)²]

where;

(x₁, y₁) is the coordinate of the first point

(x₂, y₂) is the coordinate of the second point

Now, this distance formula can also be rewritten when we square both sides to get;

(y₂ - y₁)² + (x₂ - x₁)² = d²

Thus, the missing options that makes up the distance formula between two coordinates are; (x₂ - x₁)² and d²

Read more about distance between two coordinates at; brainly.com/question/7243416

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3 years ago

Solve d=rt for t when r=75mph and d=450mi

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Step-by-step explanation:

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4 years ago

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<u>-TheUnknownScientist</u>

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3 years ago

What is the midpoint of 2,-6 and -8,5

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3 years ago

Read 2 more answers

Find the function of which is the solution of 36y"-48y'-48y=0 with initial conditions y1(0)=1. y1'(0)=0

lilavasa [31]

Answer:

$y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }$

Step-by-step explanation:

Let, $D=\frac{d}{dx}$

Now, us simplify the given differential equation and write it in terms of D,

$36y''-48y'-48y=0$

or, $3y''-4y'-4y=0$

or, $(3D^2-4D-4)y=0$

We have our auxiliary equation:

$3D^2-4D-4=0$

or, $(3D+2)(D-2)=0$

or, $D=2, - \frac{2}{3}$

Therefore our solution is,

$y=Ae^{2x}+Be^{- \frac{2}{3}x}$

and, $y'=2Ae^{2x}-\frac{2}{3}Be^{\frac{2}{3}x }$

Applying the boundary conditions, we get,

$A+B=1$

$2A-\frac{2}{3}B=0$

Solving them gives us,

$A=\frac{1}{4} ,B=\frac{3}{4}$

Hence,

$y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }$

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3 years ago