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3 years ago

29.

2 answers:

8 0

An apothem is a line drawn from the centerpoint of the polygon to one side of the polygon. There is a formula for area in terms of apothem:

A = (1/2)*(Perimeter)*(Apothem)

The perimeter of the regular hexagon is just the length of one side multiplied with the number of sides. Since a hexagon has 6 sides,

P = 6(15) = 90in

A = 1/2 * 90 * 13
A = 585 square inches

igomit [66]3 years ago

8 0

Answer:

Option 1st is correct

585 $\text{in}^2$

Step-by-step explanation:

Area(A) of a regular hexagon is given by:

$A = \frac{1}{2}P \cdot a$ ....[1]

where,

P is the perimeter and a is the apothem of the regular hexagon.

As per the statement:

An apothem 13 inches long and a side 15 inches long.

⇒a = 13 inches and side = 15 inches

Perimeter of hexagon(P) = 6s ; where s is the side

⇒P = 6(15) = 90 inches

Substitute the given values in [1] we have;

$A = \frac{1}{2} \cdot 90 \cdot 13$

Simplify:

A = 585 square inches.

Therefore, the area of a regular hexagon with an apothem 13 inches long and a side 15 inches long is, $585 in^2$

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Please show full solutions! WIll Mark Brainliest for the best answer. SERIOUS ANSWERS ONLY

Ierofanga [76]

Answer:

vertical scaling by a factor of 1/3 (compression)
reflection over the y-axis
horizontal scaling by a factor of 3 (expansion)
translation left 1 unit
translation up 3 units

Step-by-step explanation:

These are the transformations of interest:

g(x) = k·f(x) . . . . . vertical scaling (expansion) by a factor of k

g(x) = f(x) +k . . . . vertical translation by k units (upward)

g(x) = f(x/k) . . . . . horizontal expansion by a factor of k. When k < 0, the function is also reflected over the y-axis

g(x) = f(x-k) . . . . . horizontal translation to the right by k units

Here, we have ...

g(x) = 1/3f(-1/3(x+1)) +3

The vertical and horizontal transformations can be applied in either order, since neither affects the other. If we work left-to-right through the expression for g(x), we can see these transformations have been applied:

vertical scaling by a factor of 1/3 (compression) . . . 1/3f(x)
reflection over the y-axis . . . 1/3f(-x)
horizontal scaling by a factor of 3 (expansion) . . . 1/3f(-1/3x)
translation left 1 unit . . . 1/3f(-1/3(x+1))
translation up 3 units . . . 1/3f(-1/3(x+1)) +3

_____

Additional comment

The "working" is a matter of matching the form of g(x) to the forms of the different transformations. It is a pattern-matching problem.

The horizontal transformations could also be described as ...

translation right 1/3 unit . . . f(x -1/3)
reflection over y and expansion by a factor of 3 . . . f(-1/3x -1/3)

The initial translation in this scenario would be reflected to a translation left 1/3 unit, then the horizontal expansion would turn that into a translation left 1 unit, as described above. Order matters.

8 0

2 years ago

Describe the cross section

hichkok12 [17]

The answer for the exercise is the third option, which is: Hexagon.
The explanation is shown below:
As you can see in the figure attached, the cross section is a polygon of six sides and six angles. Therefore, it has six vertexes. In geometry, this type of polygon is known as "Hexagon".

6 0

3 years ago

Please Help!

enot [183]

A
Because she deposits the same amount but got 60 dollars more for the birthday money.

5 0

3 years ago

Set up an equation and solve the following problem.

nordsb [41]

Answer:

The speed of Dave is 42 miles per hour

The speed of Kent is 46 miles per hour .

Step-by-step explanation:

Given as :

The distance cover by Dave = d = 210 miles

The time taken by Dave = t hour

The speed of Dave = s miph

Again

The distance cover by Kent = D = 230 miles

The time taken by Kent = T hour

The speed of Kent = S = (s + 4 ) miph

For Dave

Time = $\dfrac{\textrm Distance}{\textrm Speed}$

So, t = $\dfrac{\textrm d miles}{\textrm s miph}$

Or, t = $\dfrac{\textrm 210 miles}{\textrm s miph}$

For Kent

Time = $\dfrac{\textrm Distance}{\textrm Speed}$

So, T = $\dfrac{\textrm D miles}{\textrm S miph}$

Or, T = $\dfrac{\textrm 230 miles}{\textrm (s + 4) miph}$

∵ Time taken by both is same

So, t = T

Or, $\dfrac{\textrm 210 miles}{\textrm s miph}$ = $\dfrac{\textrm 230 miles}{\textrm (s + 4) miph}$

Or, 210 × (s + 4) = 230 × s

Or, 210 × s + 210 × 4 = 230 × s

Or, 210 × 4 = 230 × s -210 × s

Or, 210 × 4 = 20 × s

∴ s = $\dfrac{840}{20}$

i.e s = 42 miph

So, The speed of Dave = s = 42 miles per hour

Again

The speed of Kent = S = (s + 4 ) miph

i.e S = 42 + 4

or, S = 46 miph

So, The speed of Kent = S = 46 miles per hour

Hence,The speed of Dave is 42 miles per hour

And The speed of Kent is 46 miles per hour . Answer

8 0

3 years ago

Which is a counterexample for the conditional statement shown?

Arte-miy333 [17]

Answer: First Option

The points have the same x-coordinate value.

Step-by-step explanation:

By definition, a relation is considered a function if and only if for each input value x there exists only one output value y.

So, the only way that the line that connects two points in the coordinate plane is not a function, is that these two points have the same coordinate for x.

For example, suppose you have the points (2, 5) and (2, 8) and draw a line that connects these two points.

The line will be parallel to the y axis.

Note that the value of x is the same x = 2. But when x = 2 then y = 5 and y = 8.

There are two output values (y = 8, y = 5) for the same input value x = 2.

In fact all the vertical lines parallel to the y-axis have infinite output values "y" for a single input value x. Therefore, they can not be defined as a function.

Then the correct option is:

The points have the same x-coordinate value.

7 0

3 years ago