Give an example of a rule for a pattern. list a set of numbers that fit the pattern

A cube is 6 feet on each side. What is its volume?

zlopas [31]

36ft

Explanation: 6x6=36

3 0

2 years ago

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The perimeter of a rectangle is greater than or equal to 74 meters.

Ksju [112]

Answer:

12

Step-by-step explanation:

First, multiply the length by two then, subtract the total by the sum, last divide by two.

(25 * 2)

(74 - 50)

(24/2)

Answer = 12

5 0

3 years ago

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find the surface area of a right regular hexagonal pyramid with sides 3 cm and slant heights 6cm. show all of your work.

DedPeter [7]

Answer:

The surface area of right regular hexagonal pyramid = 82.222 cm³

Step-by-step explanation:

Given as , for regular hexagonal pyramid :

The of base side = 3 cm

The slant heights = 6 cm

Now ,

The surface area of right regular hexagonal pyramid = $\frac{3\sqrt{3}}{2}\times a^{2} + 3 a \sqrt{h^{2}+ 3\times \frac{a^{2}}{4}}$

Where a is the base side

And h is the slant height

So, The surface area of right regular hexagonal pyramid = $\frac{3\sqrt{3}}{2}\times 3^{2} + 3 \times 3 \sqrt{6^{2}+ 3\times \frac{3^{2}}{4}}$

Or, The surface area of right regular hexagonal pyramid = $\frac{3\sqrt{3}}{2}\times 9 + 9 \sqrt{36+ 3\times \frac{9}{4}}$

Or, The surface area of right regular hexagonal pyramid = 23.38 + 9 × $\frac{171}{4}$

∴ The surface area of right regular hexagonal pyramid = 23.38 + 9 × 6.538

I.e The surface area of right regular hexagonal pyramid = 23.38 + 58.842

So, The surface area of right regular hexagonal pyramid = 82.222 cm³ Answer

6 0

3 years ago

Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and

zzz [600]

Answer:

$\bold{sin(38^\circ)=cos(52^\circ)}$

Step-by-step explanation:

Given that $\triangle KJL$ is a right angled triangle.

$\angle JKL = 52^\circ\\\angle KLJ = 38^\circ$

and

$\angle KJL = 90^\circ$

Kindly refer to the attached image of $\triangle KJL$ in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given $\triangle KJL$ .

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

$sin\theta =\dfrac{Perpendicular}{Hypotenuse}$

$sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}$ ....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

$cos\theta =\dfrac{Base}{Hypotenuse}$

$cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}$ ....... (2)

By (1) and (2):

sin(38°) $\neq$ cos(38°).

$cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL}$ ...... (3)

Comparing equations (1) and (3):

we get the both are same.

$\therefore \bold{sin(38^\circ)=cos(52^\circ)}$

6 0

3 years ago

The x-axis contains the base of an equilateral triangle RST. The origin is at S. Vertex T has coordinates (2h, 0) and the y-coor

frosja888 [35]

For the first part remember that an equilateral triangle is a triangle in which all three sides are equal & all three internal angles are each 60°. <span>So x-coordinate of R is in the middle of ST = (1/2)(2h-0) = h
And for the second </span><span> since this is an equilateral triangle the x coordinate of point R is equal to the coordinate of the midpoint of ST, which you figured out in the previous answer. Hope this works for you</span>

7 0

3 years ago

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