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slava [35]
3 years ago
11

Will give whoever answers correctly Brainliest

Mathematics
2 answers:
Fynjy0 [20]3 years ago
6 0

Answer:

-63

Step-by-step explanation:

because you are taking out 60 dollars and have to pay 3

PtichkaEL [24]3 years ago
5 0

Answer:

-63

Step-by-step explanation:

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The sides of the figure below measure 12, 17, and 20 centimeters. Find the
vlada-n [284]

Answer:

The perimeter should be 49 centimeters

Step-by-step explanation:

You're just adding up the centimeters that are stated in the question and then just putting the number there in the answer

8 0
2 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
Use the line plots to answer each question.
vladimir1956 [14]
Hmmmm that’s a Good question
7 0
3 years ago
I really need help with the last one if you could help it would be awesome! Tysm.
gladu [14]

it’s the second one..................

5 0
3 years ago
The Temperature started out at 92C in the morning and went down to -13C at noon. It stayed at that temparature for six hours and
ioda

the tempayure was at six because it was at 92c and went down to -13c then rose 7c so that means it was at 6 by 6 a.m.

7 0
3 years ago
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