Find the measure of Side b. Round to the nearest whole number. b = _

A square pyramid is shown sitting on its base.

allochka39001 [22]

Answer:

Area of Pyramid = 384 cm²

Step-by-step explanation:

Given

Shape: Square Pyramid

Base: 12 cm

Height: 10 cm

Required

The Surface area.

The surface area of the pyramid is calculated by

1. Calculating the area of the base square

2. Calculating the area of the triangles

3. Adding (1) and (2) above

Having highlighted these points,

Step 1: First we calculate the area of the base square

The dimension of the square is 12cm x 12cm

Let L represent the length of the square (L = 12 cm)

So, Area = L * L

Area = 12 cm * 12 cm

Area = 144 cm²

Step 2: There are four triangles in the above pyramid (because the base is a square; and a square has four sides)

The dimension of each triangle is

Base: 12 cm

Height: 10 cm

First, we calculate the area of one triangle

Area = 0.5 * base * height

Area = 0.5 * 12 cm * 10 cm

Area = 60 cm²

If the area of 1 triangle is 60 cm², the area of 4 triangles would be

Area = 4 * 60 cm²

Area = 240 cm²

Step 3: Adding (1) and (2) above

Area of Pyramid = Area of base square + Area of triangles

Area of Pyramid = 144 cm² + 240 cm²

Area of Pyramid = 384 cm²

Hence, the surface area of the pyramid is 384 cm²

5 0

3 years ago

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

2 years ago

When a force of 30 N acts on a certain object the acceleration of the object is 5m/s2 if the force is changed to 54 N what will

PtichkaEL [24]

Answer:

The correct answer is 9 m per $second^{2}$

Step-by-step explanation:

Force is defined by mass of an object times it's acceleration.

Let the mass of the object be m kilograms.

When a force of 30 N acts on a certain object the acceleration of the object is 5 m per $second^{2}$

∴ 30 = m × 5

⇒ m = 6.

If the force is changed to 54 N, let the acceleration of the object be x m per $second^{2}$ .

∴ 54 = m × x

⇒ x = $\frac{54}{6}$ = 9

Thus when the force is 54 N, the acceleration of the object is 9 m per $second^{2}$ .

6 0

4 years ago

Subtract -7x + 6 from -2x^2 + 9x - 2

JulsSmile [24]

Answer:

$( - 2 {x}^{2} + 9x - 2) - ( - 7x + 6) \\ = - 2 {x}^{2} + (9 + 7)x + ( - 2 - 6) \\ = - 2 {x}^{2} + 16x - 8$

4 0

3 years ago

The equation below describes a proportional relationship between x and y. What is the constant of proportionality?

serg [7]

Answer:

k=2/5

Step-by-step explanation:

It is the ratio of the amounts y and x:

k = y/x

Put another way: y = kx

3 0

3 years ago

Find the measure of Side b. Round to the nearest whole number. b = __ in.