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mote1985 [20]
3 years ago
7

A beach has to enclose a rectangular area, because some endangered species are nesting there. They have 200 feet of rope to rope

off the area with. What is the maximum area that they can rope off?
Mathematics
1 answer:
kkurt [141]3 years ago
6 0
Area is equal to length times width. The perimeter (the amount of rope) has to equal twice the length added to twice the width so we're left with:
A = l * w
200 = 2l + 2w
solve for either l or w
l = 100 - w
plug into the area equation to get one equation with two variables
A = w(100 - w)
A = -w^2 + 100w
take the derivative
A' = -2w + 100
set the derivative equal to zero
0 = -2w + 100
2w = 100
w = 50
This is the width that maximizes the area
with a width of 50, the length must also be 50 to have a perimeter of 200
therefore, they can rope up to 50 * 50 = 2500 ft^2
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Discount: 13%

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7 0
3 years ago
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Select a number shown by the model . Check all that apply .
dexar [7]
D,E and F are correct
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in the given figure ,o is the centr of the circle, ab is the diameter and do is perpendicular to ab. prove that angle AEC= ANGLE
maksim [4K]

The given triangles ΔDOA and ΔABC are right triangles that have a

common vertex at point <em>A</em>.

  • <u>ΔDOA is similar to ΔABC, and </u><u>∠AEC</u><u> is equal to </u><u>∠ABC,</u><u> therefore, ∠AEC = ∠ODA</u>

Reasons:

The given parameters are;

The diameter of the circle with center <em>O</em> = AB

DO ⊥ AB (DO is perpendicular to AB)

Required:

Prove that ∠AEC = ∠ODA

A two column proof is presented as follows;

Statement {}                                                 Reason

1. AB is the diameter of circle     <em> </em>{}                 1. Given

2. DO is perpendicular to AB <em> </em>{}                     2. Given

3. ∠DOA = 90° <em> </em>{}                                             3. Definition of DO ⊥ AB

4. ∠BCA = 90°  <em> </em>{}                                             4. Thales theorem

5. ∠BCA ≅ ∠BCA <em> </em>{}                                         5. Reflexive property

6. ΔDOA ~ ΔABC  <em> </em>{}                                        6. AA similarity postulate

7. ∠ABC ≅ ∠ODA  <em> </em>{}                                       7. CASTC

8. ∠ABC = ∠ODA  <em> </em>{}                                        8. Definition of congruency

9. ∠AEC ≅ ∠ABC  <em> </em>{}                                        9. Angles in the same segment

10. ∠AEC = ∠ABC  <em> </em>{}                                       10. Definition of congruency

11. ∠AEC = ∠ODA  <em> </em>{}                                        11. Transitive property of equality

In statement 6, ΔDOA is similar to ΔABC by Angle-Angle, AA, similarity

postulate, therefore, the three angles of ΔDOA are congruent to the three

angles of ΔABC.

Therefore ∠ABC ≅ ∠ODA by Corresponding Angles of Similar Triangles

are Congruent, CASTC.

Learn more about circle theorem here:

brainly.com/question/16879446

7 0
3 years ago
A jewelry store is having a 40% off sale for all necklaces. During this sale, what is the cost of a necklace that regularly cost
ELEN [110]

Answer: 18 dollars

Step-by-step explanation:

0.40 times 45 = 18 dollars

6 0
3 years ago
PLZ HELP
Alex

Answer:

The correct option is C ) 109.5°

Therefore,

m\angle E=109.5\°

Step-by-step explanation:

Given:

In Triangle DEF

d = 10

e = 18

f = 12

To Find

angle E = ?

Solution:

In Triangle DEF , Cosine Rule says

\cos E=\dfrac{f^{2}+d^{2}-e^{2}}{2fd}

Substituting the values we get

\cos E=\dfrac{12^{2}+10^{2}-18^{2}}{2\times 12\times 10}

\cos E=\dfrac{-80}{240}=-\dfrac{1}{3}

Therefore,

\angle E=\cos^{-1}(-0.3333)

m\angle E=109.5\°       ............As it is in Second Quadrant

Therefore,

m\angle E=109.5\°

4 0
4 years ago
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