Can a quadrilateral have 4 obese angles.

Answers

Answer 1

Answer:
Well, first of all, if we're talking about angles, then they can be "obtuse"
angles, but they can't be obese angles.

An obtuse angle is one that's bigger then 90° but smaller than 180°.

Aquadrilateral can have as many as 3 obtuse angles. But not 4, because
all 4 angles have to add up to 360 degrees.

So they could be like 91°, 91°, 91°, and 87° (3 obtuse angles),
or 175°, 100°, 80°, and 5° (2 obtuse angles).

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Find the midpoint of A and B where A has coordinates (2,7)
and B has coordinates (6,3).

Answers

The midpoint of A and B where A has coordinates (2,7) and B has coordinates (6,3) is (4, 5)

If B(x, y) is the midpoint of the line segment AC with end points at A(x₁, y₁) and C(x₂, y₂), the coordinates of B is:

x = (x₁ + x₂)/2; y = (y₁ + y₂)/2

Let O(x, y) be the midpoint of A and B where A has coordinates (2,7) and B has coordinates (6,3). Hence:

x = (2 + 6)/2 = 4; y = (7 + 3)/2 = 5

Hence the midpoint of A and B is (4, 5)

Find out more at: brainly.com/question/2441957

Given that the coordinates of the point A is (2,7) and the coordinates of the point B is (6,3)

We need to determine the midpoint of A and B

Midpoint of A and B:

The midpoint of A and B can be determined using the formula,

$Midpoint =((x_1+x_2)/(2), (y_1+y_2)/(2))$

Substituting the points (2,7) and (6,3) in the above formula, we get;

$Midpoint =((2+6)/(2), (7+3)/(2))$

Adding the numerator, we have;

$Midpoint =((8)/(2), (10)/(2))$

Dividing the terms, we get;

$Midpoint =(4, 5)$

Thus, the midpoint of the points A and B is (4,5)

Eleanor and Billie make 752 brownies then give them to their nine friends to share how many will they have leftover what impacted the remainder have on the answer

Answers

752/9=83 r5 your welcome

I agree with awesomepiecat. 752÷9= 83 Remainder 5.

Which is equivalent to 16 Superscript three-fourths x?RootIndex 4 StartRoot 16 EndRoot Superscript 3 x
RootIndex 4 x StartRoot 16 EndRoot cubed
RootIndex 3 StartRoot 16 EndRoot Superscript 4 x
RootIndex 3 x StartRoot 16 EndRoot Superscript 4