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3 years ago

13

What is the difference between locating 3/5 on a number line and locating √5 on a number line?

1 answer:

Alika [10]3 years ago

8 0

3/5 is 0.6 and is a rational number.

√5 however is an irrational number.

It would be quite easy to find 0.6 on a number line. It's only 6/10. It's made up of six tenths.

Now √5 is equal to 2.236067977........ to infinitum.

It's made up of many little and random fractions; to a certain degree infinitely many little and random fractions.

This means that you'd never be able to find its precise value on a number line. You could zoom in to that line for eternity and never find the precise value of √5. You'd think you're getting closer to it's precise value - but it slips away and you just can't reach it.

...Welcome to the asymptotic realm of mathematics. You've been pixelated, digitized and spaghettified. Life will never be the same again.

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3 years ago

Write an equivalent expression for 25x - 45 using the properties of operations. Hint: find the GCF of 25 and 45. You use parenth

Oksana_A [137]

Answer:

25x - 45 = 5(5x - 9)

Step-by-step explanation:

Find the greatest common factor (GCF) of 25 and 45.

You can do this several ways, but one way is to list all the factors of both numbers and find the greatest common one:

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3 years ago

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3 years ago

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dexar [7]

Answer:

$A=(72+36\sqrt{3})\ mm^2$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

Let

a ----> the height of rectangle in mm

b ---> the base of rectangle in mm

step 1

Find the base of rectangle

$b=AB+BC+CD+DE$ ----> by segment addition postulate

substitute

$b=3+3+3+3=12\ mm$

step 2

Find the height of rectangle

$a=FB+CG+GH$ ---> by segment addition postulate

substitute the given values

$a=3+CG+3\\a=6+CG$

Find the length sides CG

Applying the Pythagorean Theorem

$BG^2=BC^2+CG^2$

substitute the given values

$6^2=3^2+CG^2$

$CG^2=6^2-3^2$

$CG=\sqrt{27}\ mm$

simplify

$CG=3\sqrt{3}\ mm$

therefore

$a=(6+3\sqrt{3})\ mm$

step 3

Find the area of rectangle

$A=ab$

we have

$a=(6+3\sqrt{3})\ mm$

$b=12\ mm$

substitute

$A=(6+3\sqrt{3})(12)=(72+36\sqrt{3})\ mm^2$

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3 years ago