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umka2103 [35]
3 years ago
10

The set of the consecutive odd numbers 1, 3, 5, 7, ... , N has a sum of 400. How many numbers are in the set?

Mathematics
1 answer:
Amanda [17]3 years ago
3 0

Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.

First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.

Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:

         1, 3, 5, 7, 9, 11, 13, 15 .

Number of items in the set . . . 8
Sum of all the items in the set . . . 64

Hmmm.  That's interesting.  64 happens to be the square of 8 . 
Do you think that might be all there is to it ?

Let's check it out:

Even-numbered lists of odd numbers:

1, 3                                   Items = 2, Sum = 4
1, 3, 5, 7                           Items = 4, Sum = 16
1, 3, 5, 7, 9, 11                 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .

Amazing !  The sum is always the square of the number of items in the set !

For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.

I slogged through it on my calculator, and it's true.

I never knew this before.  It seems to be something valuable
to keep in my tool-box (and cherish always).


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d is equal to 2.

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Answer:

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8 0
3 years ago
A source of laser light sends rays AB and AC toward two opposite walls of a hall. The light rays strike the walls at points B an
Nataly [62]

Answer:

The distance between the walls is 70 m.

Step-by-step explanation:

Given: A source of laser light is at point A on the ground between two parallel walls BE and CD . The walls are perpendicular to the ground that is

BE ⊥ ED and CD ⊥ ED

AB is a ray of light which strikes the wall on the left at point B which is 30 meters above the ground. that is BE = 30 m

AC is a ray of light which strikes the wall on the right at point C. The length of AC = 80 meters.

The ray AB makes an angle of 45 degrees with the ground that is m∠BAE = 45°

The ray AC makes an angle of 60 degrees with the ground that is m∠CAD = 60°

As shown is figure attached below.

WE have to find the distance between the walls that is Length of ED

Length of ED = EA + AD

Consider the Δ AEB,

Using trigonometric ratio,

\tan\theta=\frac{perpendicular}{base}

Here \theta=45^{\circ} , perpendicular = 30 m  and base we can find.

thus,

\tan 45^{\circ}=\frac{30}{EA}

We know \tan 45^{\circ}=1

thus, EA =  30 m

Consider the Δ AEB,

Using trigonometric ratio,

\cos\theta=\frac{base}{hypotenuse}

Here \theta=60^{\circ} , hypotenuse = 80 m  and base we can find.

thus, \cos 60^{\circ}=\frac{base}{80}

We know, \cos 60^{\circ}=\frac{1}{2}

thus, Base = 40 m

AD = 40 m

Thus, the distance between the walls that is the length of ED = 30 + 40 = 70 m

4 0
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ki77a [65]
Wouldn't the answer be 12.5 because 2/4 is 0.5 and 6+6 is 12
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