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Aleks04 [339]
3 years ago
7

Which characteristics best describe an acute isosceles triangle

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
7 0

Answer:

answer is A i and iv

two sides meet at 90°

two sides are equal in lenght

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What is the slope of the line shown?
ArbitrLikvidat [17]

Answer:

2

Step-by-step explanation:

Pick two points

(2,0) and (3,2)

Then use the slope formula

m= (y2-y1)/(x2-x1)

   = (2-0)/(3-2)

   = 2/1

   =2

7 0
3 years ago
Find the dimensions of the rectangle with largest area that can be inscribed in an equilateral triangle with sides of 1 unit, if
prohojiy [21]
<span>Maximum area = sqrt(3)/8 Let's first express the width of the triangle as a function of it's height. If you draw an equilateral triangle, then a rectangle using one of the triangles edges as the base, you'll see that there's 4 regions created. They are the rectangle, a smaller equilateral triangle above the rectangle, and 2 right triangles with one leg being the height of the rectangle and the other 2 angles being 30 and 60 degrees. Let's call the short leg of that triangle b. And that makes the width of the rectangle equal to 1 minus twice b. So we have w = 1 - 2b b = h/sqrt(3) So w = 1 - 2*h/sqrt(3) The area of the rectangle is A = hw A = h(1 - 2*h/sqrt(3)) A = h*1 - h*2*h/sqrt(3) A = h - 2h^2/sqrt(3) We now have a quadratic equation where A = -2/sqrt(3), b = 1, and c=0. We can solve the problem by using a bit of calculus and calculating the first derivative, then solving for 0. But since this is a simple quadratic, we could also take advantage that a parabola is symmetrical and that the maximum value will be the midpoint between it's roots. So let's use the quadratic formula and solve it that way. The 2 roots are 0, and 1.5/sqrt(3). The midpoint is (0 + 1.5/sqrt(3))/2 = 1.5/sqrt(3) / 2 = 0.75/sqrt(3) So the desired height is 0.75/sqrt(3). Now let's calculate the width: w = 1 - 2*h/sqrt(3) w = 1 - 2* 0.75/sqrt(3) /sqrt(3) w = 1 - 2* 0.75/3 w = 1 - 1.5/3 w = 1 - 0.5 w = 0.5 The area is A = hw A = 0.75/sqrt(3) * 0.5 A = 0.375/sqrt(3) Now as I said earlier, we could use the first derivative. Let's do that as well and see what happens. A = h - 2h^2/sqrt(3) A' = 1h^0 - 4h/sqrt(3) A' = 1 - 4h/sqrt(3) Now solve for 0. A' = 1 - 4h/sqrt(3) 0 = 1 - 4h/sqrt(3) 4h/sqrt(3) = 1 4h = sqrt(3) h = sqrt(3)/4 w = 1 - 2*(sqrt(3)/4)/sqrt(3) w = 1 - 2/4 w = 1 -1/2 w = 1/2 A = wh A = 1/2 * sqrt(3)/4 A = sqrt(3)/8 And the other method got us 0.375/sqrt(3). Are they the same? Let's see. 0.375/sqrt(3) Multiply top and bottom by sqrt(3) 0.375*sqrt(3)/3 Multiply top and bottom by 8 3*sqrt(3)/24 Divide top and bottom by 3 sqrt(3)/8 Yep, they're the same. And since sqrt(3)/8 looks so much nicer than 0.375/sqrt(3), let's use that as the answer.</span>
7 0
3 years ago
Read 2 more answers
What is the range of the data set? {43.2, 46.8, 44.5, 46.8, 44.2, 41.9, 45.8, 46.9, 41.2, 46.8, 44.1}
lilavasa [31]

Answer:

the range of this answer is 5.6

3 0
3 years ago
Which description best describes the solution to the following system of equations?
sertanlavr [38]
Line y = –x + 4 intersects the line y = 3x + 3., this is the right answer
4 0
3 years ago
The regular price of a winter coat is $65.50. It is on sale for $52.40. what is the percent decrease of the winter coat
marishachu [46]
I believe the percent decrease is 20%
as when you multiply 65.5 by 20%
it gives you 13.1
when you minus 13.1 from 65.50
it gives you $52.4
6 0
3 years ago
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