3 years ago

Given AB≅CD, prove AC≅BD

Mathematics

1 answer:

Helga [31]3 years ago

6 0

It’s 56r7wjsjsjjsjwii88w9kekkekeke

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3 years ago

Complete the work to find the dimensions of the rectangle. x(x – 3) = 10 x2 – 3x = 10 x2 – 3x – 10 = 10 – 10 (x + 2)(x – 5) = 0

Norma-Jean [14]

Answer:

$Length =5$

$Width = 2$

Step-by-step explanation:

Given

$Length =x$

$Width = x -3$

$Area = 10$

$x(x-3) =10$

See comment for missing part of the question

Required

Complete the expression to determine the dimension of a rectangle

We have:

$x(x-3) =10$

Open bracket

$x^2 -3x = 10$

Equate to 0

$x^2 -3x - 10 =0$

Expand

$x^2 + 2x - 5x - 10 = 0$

Factorize

$x(x + 2) - 5(x + 2) = 0$

Factor out x + 2

$(x - 5)(x + 2) = 0$

Solve for x

$x - 5 = 0$ or $x + 2 = 0$

$x = 5$ or $x = -2$

The value of x cannot be negative

So:

$x = 5$

Recall that:

$Length = x$

$Width = x - 3$

So:

$Length =5$

$Width = 2$ ---- i.e. 5 - 3

6 0

2 years ago