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torisob [31]
3 years ago
15

Suppose a seventh graders birthday is today, and she is 12 years old. How old was she 3 1/2 years ago? Write an equation, and us

e a number line to model your answer
Mathematics
1 answer:
german3 years ago
4 0

Answer:

he age of girl 3 \frac{1}{2} years ago is 8.5 years  .

Step-by-step explanation:

Given as :

The present age of seventh grader girl = 12 years

Let The age of her 3 \frac{1}{2} years ago = x years

So The age of girl \frac{7}{2} years ago = x years

<u>Now, According to question</u>

The age of girl 3.5 years ago = 12 - 3.5

Or,  The age of girl 3.5 years ago = 8.5 years

Hence The age of girl 3 \frac{1}{2} years ago is 8.5 years  . Answer

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What's the approximate area of a segment of a circle with a height 6 m and the length of the chord is 20 m? Round your answer to
yuradex [85]
Hello!

The Correct Answer to this would be 100%:

Option "85.4".

(Work Below)

Given:
height = 6m
chord = 20 m

We need to find the radius of the circle.

20 m = 2 √ [ 6m( 2 x radius - 6 m ) ] 
20 m / 2 = 2 √[ 6m( 2 x radius - 6 m ) ] / 2 
10 m = √ [ 6m( 2 x radius - 6 m ) ] 
(10 m)² = √[ 6m( 2 x radius - 6 m ) ] ² 
100 m² = 6 m( 2 x radius - 6 m ) 
100 m² = 12 m x radius - 36 sq m 
100 m² + 36 m² = 12 m x radius - 36 m² + 36 m² 
136 m² = 12 m x radius 
136 m² / 12 m = 12 m x radius / 12 m 
<span>11.333 m = radius 
</span>
the area beneath an arc: 

Area = r² x arc cosine [ ( r - h ) / r ] - ( r - h ) x √( 2 x r x h - h²<span> ). 
</span>
r² = (11.333 m)² = 128.444 m² 
r - h= 11.333 m - 6 m = 5.333 m 
r * h = 11.333 m x 6 m = 68 m²

Area = 128.444 m² x arc cosine [ 5.333 m / 11.333 m ] - 5.333 m x √[ 2 x 68 m² - 36 m² ] 

Area = 128.444 m² x arc cosine [ 0.4706 ] - 5.333 m x √ [ 100m² ] 

Area = 128.444 m² x 1.0808 radians - 5.333 m x 10 m 

Area = 138.828 m² - 53.333 m² 

Area = 85.4 m<span>²
</span>

Hope this Helps! Have A Wonderful Day! :)


And as Always...

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