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Bess [88]
2 years ago
8

A square pyramid and a cube have the same total surface area. The width of the cube is 8 centimeters. The base of the pyramid is

equal to base of the cube. What is the slant height of the square pyramid?
Mathematics
1 answer:
azamat2 years ago
6 0

Answer:

Step-by-step explanation:

surface area of cube = 6×8² = 384 cm²

lateral area of pyramid = surface area of pyramid - base area

= 384 - 8²

= 320 cm²

area of triangular face = 320/4 = 80 = ½(base)(slant height) = ½(8)(slant height)

slant height = 20 cm

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Outline the process for inscribing a regular hexagon in a circle. Perform the construction in GeoGebra
mr_godi [17]

The following are the general steps to draw a hexagon inscribed in a circle:

First draw a circle with center A through B:

Then the intersection points will be vertex C and vertex D of the hexagon. Doing this for each vertex (from left to right) we will get the following:

4 0
1 year ago
Which two equations would be most appropriately solved by using the zero product property? Select each correct answer.
LekaFEV [45]
The zero product property tells us that if the product of two or more factors is zero, then each one of these factors CAN be zero.

For more context let's look at the first equation in the problem that we can apply this to: (x-3)(x+4)=0

Through zero property we know that the factor (x-3) can be equal to zero as well as (x+4). This is because, even if only one of them is zero, the product will immediately be zero.

The zero product property is best applied to factorable quadratic equations in this case.

Another factorable equation would be 2x^{2}+6x=0 since we can factor out 2x and end up with 2x(x+3)=0. Now we'll end up with two factors, 2x and (x+3), which we can apply the zero product property to.

The rest of the options are not factorable thus the zero product property won't apply to them.
3 0
3 years ago
Read 2 more answers
Consider the trianglular region with vertices (0,0) (6,0) and (0,6). find the x and y coordinates of the centroid
kondor19780726 [428]
Let A=(0,0)(x₁,x₂), B=(6,0)(x₂,y₂) and C=(0,6)(x₃,y₃)

Centroid of ΔABC is given by,

G(x,y) = [x₁+x₂+x₃/3 , y₁+y₂+y₃/3] = [0+6+0/3 , 0+0+6/3] = [2,2]
6 0
2 years ago
72
Zigmanuir [339]

Answer:

Ai. Arithmetic sequence

Aii. Tn = 5 + 7n

Bi. Geometric

Bii. Tn = 8 × 2ⁿ¯¹

Step-by-step explanation:

To successfully answer the questions given above, note the following:

1. If the sequence is Arithmetic, then:

2nd – 1st = 3rd – 2nd = common difference (d)

2. If the sequence is geometric, then,

2nd / 1st = 3rd / 2nd = common ratio (r)

3. A sequence can not be arithmetic geometric at the same time.

4. The nth term of arithmetic sequence is:

Tn = a + (n – 1)d

5. The nth term of geometric sequence is:

Tn = arⁿ¯¹

A. Sequence => 12, 19, 26

i. Determination of the type of sequence.

We'll begin by calculating the common difference

1st term = 12

2nd term = 19

3rd term = 26

Common difference (d) = 2nd – 1st

d = 19 – 12 = 7

OR

d = 3rd – 2nd

d = 26 – 19 = 7

Since a common difference exist in the sequence, the sequence is arithmetic sequence.

ii. Determination of the nth term.

Common difference (d) = 7

1st term (a) = 12

nth term (Tn) =?

Tn = a + (n – 1)d

Tn = 12 + (n – 1)7

Tn = 12 + 7n – 7

Tn = 5 + 7n

B. Sequence => 8, 16, 32

Bi. Determination of the type of sequence.

Let us begin by calculating the common ratio.

1st term = 8

2nd term = 16

3rd term = 32

Common ratio (r) = 2nd / 1st

r = 16 / 8

r = 2

OR

r = 3rd / 2nd

r = 32 / 16

r = 2

Since a common ratio exist in the sequence, the sequence is geometric.

Bii. Determination of the nth term.

Common ratio(r) = 2

1st term (a) = 8

nth term =?

Tn = arⁿ¯¹

Tn = 8 × 2ⁿ¯¹

8 0
2 years ago
Find the ​slope-intercept form of the equation of the line that passes through the point P and makes angle with the positive ​x-
SIZIF [17.4K]

Answer:

The answer is "y= -x+7"

Step-by-step explanation:

If \bold{\theta = 135^{\circ}}

point: p=(2,5)

Formula:

\to m= \tan \theta \\\\\to (y-y_1)= m(x-x_1)

\to m= \tan \ 135^{\circ} =  -1\\\\\to x_1= 2\\\\\to y_1=5

Put the value in the above formula:

\to y- 5 = -1(x-2)\\\\\to y- 5 = -x+2\\\\\to x+y = 5+2\\\\\to x+y=7

\to y= -x+7

3 0
2 years ago
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