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galina1969 [7]
3 years ago
12

An archery target consists of a circular bull's-eye with radius x, surrounded by four rings with width y. What is the area of th

e outermost ring in terms of x and y?
Mathematics
1 answer:
amid [387]3 years ago
4 0
<span>given:
   bull's eye radius= x
 width of surrounding rings=y

   solution:
   Radius of the circle=x+4y
Area of the outermost ring=Area of the circle-Area of the penultimate ring =Ď€(x+4y)^2-Ď€(x+3y)^2
=Ď€(x^2+8xy+16y^2-x^2-9y^2-6xy)
 =Ď€(2xy+7y^2)
 hence the area of the outermost ring in terms of x and y is Ď€(2xy+7y^2).</span>
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Alenkinab [10]

Answer:

Option B and D are correct.

Step-by-step explanation:

Given: A line passes through the points (2,4) and (5,6).

* Case 1:

If a line passes through the points (2, 4) and (5, 6)

Point slope intercept form:

for any two points (x_1,y_1) and (x_2, y_2)

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m =\frac{y_2-y_1}{x_2-x_1}

First calculate slope for the points (2, 4) and (5, 6);

m = \frac{y_2-y_1}{x_2-x_1} =\frac{6-4}{5-2} = \frac{2}{3}

then, by point slope intercept form;

y-4=\frac{2}{3}(x-2)

* Case 2:

If a line passes through the points (5, 6) and (2, 4)

First calculate slope for the points (5, 6) and (2, 4);

m = \frac{y_2-y_1}{x_2-x_1} =\frac{4-6}{2-5} = \frac{-2}{-3}= \frac{2}{3}

then, by point slope intercept form;

y-6=\frac{2}{3}(x-5)

Yes, the only equation of line from the given options  which describes the given line are;

y-4=\frac{2}{3}(x-2)  and y-6=\frac{2}{3}(x-5)



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3 years ago
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56.6 - 3.2 = 53.4
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But, let's say you get a haircut+shampoo

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6 0
3 years ago
a child should no longer play in the kid zone area when they reach 4'4" tall joe is 54" tall is he allowed to play in the kid zo
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