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UkoKoshka [18]
3 years ago
7

A circle has radius of 18 cm. Find the area of the smaller of the two regions determined by a chord with length of 18√2 cm . Hin

t: (Look for right triangles using Pythagorean Theorem Converse)
Mathematics
2 answers:
boyakko [2]3 years ago
8 0
The area is 81*pi-162. Look at the triangle formed by the chord and the two radii connected to the endpoints of the chord. Since 18^2+18^2=(18*\sqrt2)^2, this is an isosceles right triangle by Pythagorean theorem. The area of the quarter circle containing the region we look for is pi*18^2/4=81*pi. The area of the right triangle is 18*18/2=162. The difference between these two areas, or 81*pi-162, is the area of the smaller region.
madreJ [45]3 years ago
3 0

Answer:

81pi - 162

Step-by-step explanation:

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Please look at this multiple choice, thanks!
Nastasia [14]

Step-by-step explanation:

\frac{y - y1}{x - x1}  =  \frac{y2 - y1}{x2 - x1}

\frac{y - 0}{x - ( - 1)}  =  \frac{3 - 0}{1 - ( - 1)}

\frac{y}{x + 1}  =  \frac{3}{1 + 1}

\frac{y}{x + 1}  =  \frac{3}{2}

y =  \frac{3}{2} (x + 1)

y =  \frac{3}{2} x +  \frac{3}{2}

option B

7 0
3 years ago
In 1928, when the high jump was first introduced as a women's sport at the Olympic Games, the winning jump for women was 70.0 in
sveta [45]

Answer: year 2194



Explanation:



1) Model the jump of the women:


i) initial jump = 70.0 in


ii) increase rate = 0.48% per year


iii) equation: h = 70.0(1 +0.0048)ⁿ


2) Model the jump of the men:


i) initial jump 86.5 inches.


ii) increase rate, 0.4%.


iii) equation: h = 86.5 (1 + 0.004)ⁿ


3) Equal the two equations (h = h) to find when the jumps will be equal:


i) 70.0(1 +0.0048)ⁿ = 86.5 (1 + 0.004)ⁿ


ii) 70.0(1.0048)ⁿ = 86.5 (1.004)ⁿ


ii) [1.0048ⁿ / 1.004ⁿ] = 86.5/70.0


iii) [1.0048/ 1.0040]ⁿ = 86.5/70.0


iv) n log (1.0048 / 1.0040) = log ( 86.5/70)


v) n = log (86.5/70) / log (1.0048/1.0040) ≈ 266


3) year = 1928 + 266 = 2194


That is in the year 2194

8 0
3 years ago
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What is the domain and range please?
Vinvika [58]

Answer:

domain is the max and min x values

range is the max and min y values

Step-by-step explanation:

8 0
3 years ago
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How many gallons of paint would you need to paint a dog house that is 372ft in total?
Shalnov [3]
It would be 44gallons
6 0
3 years ago
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What are the vertical and horizontal asymptotes for the function f(x)=<br> 3x2/x2-4
Alecsey [184]

Answer:  f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.

Step-by-step explanation:

Given function: f(x)=\dfrac{3x^2}{x^2-4}

The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.

i.e. x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2

Hence, f(x) will have vertical asymptotes at x=-2 and x=2.

To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.

So, the graph will horizontal asymptote at y=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}

i.e. y=\dfrac{3}{1}=3

Hence, f(x) will have horizontal asymptote at y=3.

3 0
3 years ago
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