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3 years ago

Figure ABCD is graphed on a coordinate plane.

Mathematics

2 answers:

Lesechka [4]3 years ago

6 0

20.9 Is your answer

Phoenix [80]3 years ago

3 0

1. BC is 1+3=4 (units)

2. AD is 3+5= 8 (units)

3. Draw the altitude CH from point C, as shown in the figure.

4. Then triangle CHD is a right triangle, with hypothenuse CD, and sides HD and CH.

5. $CD^2=HD^2+CH^2$

$CD^2=2^2+4^2=4+16=20$

$CD= \sqrt{20}= \sqrt{4*5}= \sqrt{4} * \sqrt{5}=2 \sqrt{5}= 4.47$

6. AB=CD=4.47 because the trapezoid is isosceles.

7. P= BC+AD+AB+CD=4+8+4.47+4.47=20.9 (units)

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3 years ago

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1 year ago

Three people whowork full time are working together on a project but their totaltime on the project is to be equivalent to that

Art [367]

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2 years ago

Plzzzzz help will give brain and points!!! Explain why a quadratic equation with a positive discriminant has two real solutions,

NISA [10]

Answer:

The given equation is -2x^2 = -8x + 8

This can be rewritten as 2x^2 - 8x + 8 = 0

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5 0

3 years ago

The answer choices are 11 1 2.48 6

Fudgin [204]

Answer:

The ball reached its maximum height of ( $11\ yards$ ) in ( $1\ second$ ).

Step-by-step explanation:

This question is essentially asking one to find the vertex of the parabola formed by the given equation. One could plot the equation, but it would be far more efficient to complete the square. Completing the square of an equation is a process by which a person converts the equation of a parabola from standard form to vertex form.

The first step in completing the square is to group the quadratic and linear term:

$h(t)=-5t^2 + 10t + 6\\\\h(t) = (-5t^2 + 10t) + 6$

Now factor out the coefficient of the quadratic term:

$h(t)=-5(t^2 -2t) + 6$

After doing so, add a constant such that the terms inside the parenthesis form a perfect square, don't forget to balance the equation by adding the inverse of the added constant term:

$h(t) = -5(t^2 -2t) + 6\\\\h(t) = -5(t^2 -2t + 1 -1 ) + 6$

Now take the balancing term out of the parenthesis:

$\\\\h(t)=-5(t^2 -2t + 1) + 6 + ((-1)(-5))$

Simplify:

$h(t) = -5(t^2 -2t + 1) + 6 + 5\\\\h(t) = -5(t-1)^2 + 11$

The x-coordinate of the vertex of the parabola is equal to the additive inverse of the numerical part of the quadratic term. The y-coordinate of the vertex is the constant term outside of the parenthesis. Thus, the vertex of the parabola is:

$(1, 11)$

8 0

2 years ago