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mezya [45]
3 years ago
7

Find the area of a regular hexagon with an apothem 11.4 yards long and a side 13 yards long. Round your answer to the nearest te

nth.
Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

Step-by-step explanation:

area=6×11.4×13÷2=444.6 yd²

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HELP ASAP Linear Equations
Scilla [17]

Answer:

4th choice: I and IV only

Step-by-step explanation:

Change the form of the given equation to standard form.

y = -2/3 x + 2

3y = -2x + 6

2x + 3y = 6

Now compare this equation with the 4 choices.

Any equation that has the same x- and y-terms but a different constant term is parallel.

I. is parallel.

II. and III. are not parallel.

Look at IV.:

4x + 6y = 3

Divide both sides by 2:

2x + 3y = 3/2

IV. is also parallel.

Answer: 4th choice: I and IV only

4 0
3 years ago
Read 2 more answers
What is the 6th term of the geometric sequence f(1) = 12, f(n) = -2/3f(n-1)
Troyanec [42]
\bf \begin{array}{ccll}
\stackrel{n^{th}}{term}&value\\\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\\\
1&12\\\\
2&-\frac{2}{3}f(2-1)\\\\
&-\frac{2}{3}f(1)\\\\
&-\frac{2}{3}(12)\\\\
&-8\\\\
3&-\frac{2}{3}f(3-1)\\\\
&-\frac{2}{3}f(2)\\\\
&-\frac{2}{3}(-8)\\\\
&\frac{16}{3}
\end{array}

\bf \begin{array}{cccl}
\qquad &\qquad \\
4&-\frac{2}{3}f(4-1)\\\\
&-\frac{2}{3}f(3)\\\\
&-\frac{2}{3}\left( \frac{16}{3} \right)\\\\
&-\frac{32}{9}\\\\
5&-\frac{2}{3}f(5-1)\\\\
&-\frac{2}{3}f(4)\\\\
&-\frac{2}{3}\left( -\frac{32}{9} \right)\\\\
&\frac{64}{27}\\\\
6&-\frac{2}{3}f(6-1)\\\\
&-\frac{2}{3}f(5)\\\\
&-\frac{2}{3}\left(  \frac{64}{27}\right)
\end{array}
5 0
3 years ago
Plz help I need this by tomorrow ​
Rufina [12.5K]

Ik this doesn't help but maybe look the questions up online bc those are even confusing for me to understand. Really sorry I couldnt help :(

6 0
3 years ago
Read 2 more answers
Please help me on this question :(
Anni [7]
Graph A is the correct answer
Good luck
3 0
3 years ago
Can you match it??. Only if you know please
Dennis_Churaev [7]

Answer:

Step-by-step explanation:

triangle: a+b>c. If two lengths are put together, and equals more than the length of the third side, then it's a triangle

right triangle: a2+b2=c2 if the square of the two lengths equal the square of the third, then it's a right triangle

acute triangle: a2+b2>c2 if the square of the two lengths are greater than the square of the third it's acute

obtuse triangle:a2+b2<c2 if the square of the two lengths are less than the square of the third it's obtuse

8 0
3 years ago
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