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AysviL [449]
3 years ago
12

A quadrilateral has diagonals that are congruent and bisect opposite pairs of

Mathematics
1 answer:
gtnhenbr [62]3 years ago
8 0
If the diagonals of a quadrilateral are perpendicular, then it is a rhombus. False, diagonals don't have to be congruent or bisect each other. The diagonals of a rectangle bisect its angles. A kite with all consecutive angles congruent must be a square.
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Can i have help please. Step by step
nignag [31]

Answer: It's in the step-by step explanation

Step-by-step explanation:

I just learned about this too. I'll use what I know to help you out.

According to whatever law of the circle, where you have two intersecting lines within the bounds of a circle(that'd be TQ and SW), the product of the divided segments will equal each other.

So to put that in terms, TU times QU = SU times WU.

So let's get the value of segment TU, which is 1.5

Then let's get the value of segment of QU, which is 4.

Now let's get the value of WU, which is 3. We don't know what SU is yet. So put it in algebraic form.

1.5(4)=3x

6=3x

2=x

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0.729 rounded by the nearest tenth
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If α, β are the zeroes of the polynomials f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) =
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Answer:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

Step-by-step f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)explanation:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

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3 years ago
Find the solution to Y=-x^2+3 for X=-3,0, and 3
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Answer:

3-,3,0,3

Step-by-step explanation:

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