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kozerog [31]
2 years ago
14

A theater has 50 seats on the front row. There are four additional seats in each following row. PART B: How many seats would the

re be in the 20th row?
Mathematics
1 answer:
Roman55 [17]2 years ago
7 0
132, if you add four to 50, and keep going until you get to 20
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Find the perimeter of Triangle ABC with the verticies A(-5,5) B(3,-5) and C(-5,1
finlep [7]

Hello!
 
Let's calculate a distance between two points using the Pythagorean theorem:<span>

</span>d ^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2

Data:

x_B = 3 &#10;&#10;x_A = -5 &#10;&#10;y_B = -5 &#10;&#10;y_A = 5 &#10;&#10;d_ {AB} =?

Solving: distance from A to B

d^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2 &#10;&#10;d^ 2_ {AB} = (3 - (-5)) ^ 2 + (-5 - 5) ^ 2 &#10;&#10;d^ 2_ {AB} = (8) ^ 2 + (-10) ^ 2 &#10;&#10;d^ 2_ {AB} = 64 + 100 &#10;&#10;d^ 2_ {AB} = 164 &#10;&#10;d_{AB} = \sqrt {164} &#10;&#10;d_{AB} = \sqrt {2 ^ 2 * 41} &#10;&#10;\boxed {d_ {AB} = 2 \sqrt {41}}

<span>Solving:
 
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d ^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2

Data:

x_C = -5&#10; &#10;x_A = -5&#10; &#10;y_C = 1 &#10;&#10;y_A = 5 &#10;&#10;d_ {AC} =?


d^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2 &#10;&#10;d^ 2_ {AC} = (-5 - (-5)) ^ 2 + (1 - 5) ^ 2 &#10;&#10;d^ 2_ {AC} = (0) ^ 2 + (-4) ^ 2  &#10;&#10;d^ 2_ {AC} = 0 + 16  &#10;&#10;d^ 2_ {AC} = 16 &#10;&#10;d_ {AC} = \sqrt {16} &#10;&#10;\boxed {d_ {AC} = 4}&#10;<span>
</span>

Solving:

distance from B to C

d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2

Data:

x_C = -5 &#10;&#10;x_B = 3 &#10;&#10;y_C = 1 &#10;&#10;y_B = -5 &#10;&#10;d_ {BC} =?


d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2&#10; &#10;d^ 2_ {BC} = (-5 - 3) ^ 2 + (1 - (-5)) ^ 2&#10;&#10;d^ 2_ {BC} = (-8) ^ 2 + (6) ^ 2  &#10;&#10;d^ 2_ {BC} = 64 + 36&#10;&#10;d^ 2_ {BC} = 100 &#10;&#10;d_ {BC} = \sqrt {100}&#10;&#10;\boxed {d_ {BC} = 10}

<span>Now let's calculate the perimeter of the triangle ABC, knowing that the perimeter is the sum of the sides, then:
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p = d_ {AB} + d_ {AC} + d_ {BC}  &#10;&#10;

p = 2\sqrt {41} + 4 + 10

&#10;\boxed {\boxed {p = 14 + 2 \sqrt {41}}} \end {array}} \qquad \quad \checkmark
8 0
3 years ago
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2 years ago
How would i solve this the shaded region is the whole inside circle.​
rosijanka [135]

Answer:

34.48cm²

Step-by-step explanation:

Assuming the shaded area doesn't contain the triangle:

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area of triangle = 4(2)/2

area = 4

area of circle = πr²

area = π3.5²

area = 38.48

area of shaded = 38.48 - 4

area = 34.48cm²

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16 peaches in each box
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