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andriy [413]
2 years ago
12

Which is the hypotenuse-angle theorem? If the hypotenuse and an acute angle of a right triangle are congruent to the correspondi

ng parts of another right triangle, then the triangles are congruent. If the hypotenuse and an obtuse angle of a right triangle are congruent to the corresponding parts of another right triangle, then the triangles are congruent. If the hypotenuse and an acute angle of a right triangle are congruent to the corresponding parts of another right triangle, then the triangles are complimentary. If the hypotenuse and an acute angle of a right triangle are congruent to the corresponding parts of another right triangle, then the triangles are supplementary.
Mathematics
1 answer:
Rufina [12.5K]2 years ago
8 0
i would say this is the third one
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The answer is D complete the square
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6 times a number is 27 less than the square of that number. Find the positive solution.
Verdich [7]

Let x = your number.

6x = x^2 - 27. Subtract 6x from each side.

0 = x^2 - 6x - 27. Factor

0 = (x-9) (x+3). Set each term equal to zero

(x+3) = 0. Subtract 3 from each side.

x = -3. This is the negative solution.

(x-9) = 0. Add 9 to each side.

x = 9. This is the positive solution.

6 0
3 years ago
Three angles of a hexagon are congruent. The other three angles are also congruent, each with a measure twice that of the first
Serga [27]
The sum of the interior angles of a hexagon is 720 degrees. If 3 angles are congruent, each with measure x degrees, and the other 3 angles are also congruent to each other, each with measure 2x degrees, then the total sum of all 6 angles would be 3x + 3(2x) = 9x. If this is equal to 720, then x = 80 degrees, while 2x = 160 degrees.
Therefore, there are 3 80-degree angles and 3 160-degree angles.
3 0
3 years ago
If 1200 cm^2 of material is available to make a box with a square base and an open top find the largest possible volume of the b
strojnjashka [21]
Surface area of box=1200 cm² 
<span>Volume of box=s²h </span>
<span>s = side of square base </span>
<span>h = height of box </span>
<span>S.A. = s² + 4sh </span>
<span>S.A. = surface area or 1200 cm², s²
 = the square base, and 4sh = the four 'walls' of the box. </span>
<span>1200 = s² + 4sh </span>
<span>1200 - s² = 4sh </span>
<span>(1200 - s²)/(4s) = h </span>


<span>v(s) = s²((1200 - s²)/(4s)) </span>
<span>v(s) = s(1200 - s²)/4 . </span>
<span>v(s) = 300s - (1/4)s^3</span>

by derivating
<span>v'(s) = 300 - (3/4)s² </span>
<span>0 = 300 - (3/4)s² </span>
<span>-300 = (-3/4)s² </span>
<span>400 = s² </span>
<span>s = -20 and 20. </span>
again derivating
<span>v"(s) = -(3/2)s </span>
<span>v"(-20) = -(3/2)(-20) </span>
<span>v"(-20) = 30 </span>
<span>v"(20) = -(3/2)(20) </span>
<span>v"(20) = -30 </span>


<span>v(s) = 300s - (1/4)s^3 </span>
<span>v(s) = 300(20) - (1/4)(20)^3 </span>
<span>v(s) = 6000 - (1/4)(8000) </span>
<span>v = 6000 - 2000 
v=4000</span>
4 0
2 years ago
What is the answer? I am so lost- thanks for any help given.
aev [14]

Answer:

i think that the answer is 13 meters

Step-by-step explanation:

7 0
3 years ago
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