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3 years ago

8

Find the area of an equilateral triangle (regular 3 Gon) with the given measurement 9 in perimeter a=SQ.in

1 answer:

blsea [12.9K]3 years ago

4 0

Answer:

3.897 $in^{2}$

Step-by-step explanation:

equilateral triangles are also equiangular, meaning the have equal angles.

Triangle sum theory says that angles of a triangle add up to 180.

That means each angle is 60.

A = bh/2

You need the (h). The base of is 3. Perimeter = 9, so each side is 3

Draw a perpendicular line for the height. The line cuts the base in half (1.5)

Using trigonometry you can find the height.

tan 60° = h/1.5

h = height, 1.5 is half of 3, 60° is the base angle.

multiply each side by 1.5

1.5(tan 60°) = h

h=2.598

then substitute h into formula

A= <u>(2.598)(3) </u>

2

A = 3.897 rounded

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expeople1 [14]

Answer:

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3 years ago

pantera1 [17]

Answer:

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3 years ago

Suppose that the functions u and w are defined as follows.

jasenka [17]

Answer(there are assumptions for this answer that you need to confirm and look at):

Assumptions: $u(x)=x^2+3$ and $w(x)=\sqrt{x+2}$

Answer if the operation is multiplication:

If you meant a closed dot which is the symbol for multiplication.

$(u \cdot w)(2)=14$

$(w \cdot u)(2)=14$

Answer if the operation is composition:

If you meant an open dot which is the symbol for composition.

$(u \circ w)(2)=7$

$(w \circ u)(2)=3$

Note: I don't know if you actually meant $w(x)=\sqrt{x+2}$ or if $w(x)=\sqrt{x}+2$ . Please let me know one way or the other.

Step-by-step explanation:

If we assume the functions are:

$u(x)=x^2+3$

$w(x)=\sqrt{x+2}$

$u \cdot w=w \cdot u$ since multiplication is commutative.

$u(2)=2^2+3$

$u(2)=4+3$

$u(2)=7$

$w(2)=\sqrt{2+2}$

$w(2)=\sqrt{4}$

$w(2)=2$

We are asked to find $(u \cdot w)(2)$ and $(w \cdot u)(2)$ .

The order doesn't matter in multiplication.

$(u \cdot w)(2)$

$u(2) \cdot w(2)$

$7 \cdot 2$

$14$

$(w \cdot u)(2)$

$w(2) \cdot u(2)$

$2 \cdot 7$

$14$

Now you might have meant composition which symbolized with an open circle, not a closed one.

$(u \circ w)(2)$

$u(w(2))$

$u(2)$ since $w(2)=2$

$2^2+3$

$4+3$

$7$

$(w \circ u)(2)$

$w(u(2))$

$w(7)$ since $u(2)=7$

$\sqrt{7+2}$

$\sqrt{9}$

$3$

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4 years ago

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3 years ago

Solnce55 [7]

............15n+12b

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3 years ago