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poizon [28]
3 years ago
10

Find the area of an equilateral triangle (regular 3-gon) with the given measurement in terms with a radical. 6-inch side. A = sq

. in.
Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0
Draw a straight line from the mid point of one side to the opposite vertex. Use the Pythagorean Theorem to find the length of this line. Let b = this line (the height). a^2 + b^2 = c^2 3^2 + b^2 = 6^2 9 + b^2 = 36 b^2 = 27 b = 5.2 Now use this formula to find the area of this triangle. A = 1/2bh
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Geometry review: Which term can be defined as lines in the same plane
BARSIC [14]

Answer:

Parallel lines are lines in the same plane

that never intersect and are always the same distance apart

6 0
3 years ago
Read 2 more answers
Using the digits 1 to 9, at most one time each, fill in the boxes so that the points make a parallelogram. Mathematically, use p
Lera25 [3.4K]

Answer:

\left\begin{array}{ccc}A\left( \boxed{3}  , \boxed{8} \right)&B\left( \boxed{9}  , \boxed{6} \right )\\\\C\left( \boxed{1}  ,  \boxed{4} \right)&D\left( \boxed{7}  ,  \boxed{2} \right )\end{array}\right

Step-by-step explanation:

The coordinates are:

\left\begin{array}{ccc}A\left( \boxed{3}  , \boxed{8} \right)&B\left( \boxed{9}  , \boxed{6} \right )\\\\C\left( \boxed{1}  ,  \boxed{4} \right)&D\left( \boxed{7}  ,  \boxed{2} \right )\end{array}\right

The parallelogram is attached below.

To verify if these coordinates form a parallelogram, we show that:

  • AB=CD; and
  • AC=BD

Using Distance Formula

AB = \sqrt{(6-8)^2+(9-3)^2} = \sqrt{(-2)^2+(6)^2} = \sqrt{40} $ units

CD = \sqrt{(2-4)^2+(7-1)^2} = \sqrt{(-2)^2+(6)^2} = \sqrt{40} $ units

AC = \sqrt{(8-4)^2+(3-1)^2} = \sqrt{(4)^2+(2)^2} = \sqrt{20} $ units

BD = \sqrt{(6-2)^2+(9-7)^2} = \sqrt{(4)^2+(2)^2} = \sqrt{20} $ units

Since AB=CD; and AC=BD, the coordinates A, B, C, and D form the vertex of a parallelogram.

3 0
4 years ago
Plz plz answer this i'll give 11 points
shutvik [7]

Answer:

A =bh

Step-by-step explanation:

2 1/2 times 5 1/4 = 13.125

4 0
3 years ago
If a polynomial function f(x) has roots 1+square root of 2 and-3, what must be a factor of f(x)?
nevsk [136]
To find the factor a a polynomial from its roots, we are going to seat each one of the roots equal tox, and then we are going to factor backwards.

We know for our problem that one of the roots of our polynomial is -3, so lets set -3 equal tox and factor backwards:
x=-3
x+3=0
(x+3) is a factor of our polynomial.

We also know that another root of our polynomial is 1+ \sqrt{2}, so lets set 1+ \sqrt{2} equal to x and factor backwards:
x=1+ \sqrt{2}
x-1= \sqrt{2}
x-1- \sqrt{2}=0
(x-(1+ \sqrt{2})=0
((x-(1+ \sqrt{2} )) is a factor of our polynomial.

We can conclude that there is no correct answer in your given choices.
8 0
4 years ago
Solve for y: y/3=-4+y<br><br> What is y?
Dmitry [639]

Answer:

y = 6

Step-by-step explanation:

6 0
3 years ago
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