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Alexus [3.1K]
2 years ago
8

Polygon ABCD has vertices A(1,3), B(1,6), C(4,6), and D(5,2)

Mathematics
1 answer:
Galina-37 [17]2 years ago
7 0

Answer: A (0.5, 1.5)   B(0.5, 3)    C(2, 3)    D(2.5, 1)

Step-by-step explanation:

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(x^4)^2 <br> how do I simplify this expression?
castortr0y [4]

Answer:

x^8

Step-by-step explanation:

(x^4)^2

~Apply power rule [ (a^b)^c = a^bc ]

x^4(2)

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x^8

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Which expression is equivalent to the following complex fraction?
I am Lyosha [343]

what a mess?!

please write your question better...

if you don't want the wrong answer

Step-by-step explanation:

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3 years ago
Line segment 19 units long running from (x,0) ti (0, y) show the area of the triangle enclosed by the segment is largest when x=
Debora [2.8K]
The area of the triangle is

A = (xy)/2

Also,

sqrt(x^2 + y^2) = 19

We solve this for y.

x^2 + y^2 = 361

y^2 = 361 - x^2

y = sqrt(361 - x^2)

Now we substitute this expression for y in the area equation.

A = (1/2)(x)(sqrt(361 - x^2))

A = (1/2)(x)(361 - x^2)^(1/2)

We take the derivative of A with respect to x.

dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]

dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]

Now we set the derivative equal to zero.

(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0

-2x^2 + 361 = 0

-2x^2 = -361

2x^2 = 361

x^2 = 361/2

x = 19/sqrt(2)

x^2 + y^2 = 361

(19/sqrt(2))^2 + y^2 = 361

361/2 + y^2 = 361

y^2 = 361/2

y = 19/sqrt(2)

We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
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3 years ago
Ramon travels 20 miles in 20 minutes, how many miles can he travel in 120 minutes?
Elis [28]

Answer: 120

Step-by-step explanation:

20 divided by 20 is 1.

We have to multiply this by 120.

120 x 1 = 120. It is the answer.

4 0
3 years ago
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