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3 years ago

I need help with this question please

Mathematics

2 answers:

Neporo4naja [7]3 years ago

6 0

Given:

Number of faces = 6

Number of vertices = 8

Formula used:

Euler's formula:

F + V - E = 2

F = Number of Faces of Polyhedron

V = Number of Vertices of Polyhedron

E = Number of Edges of Polyhedron

Calculations:

Let the polyhedron have E edges,

6 + 8 - E = 2

14 - E = 2

E = 14 - 2

E = 12

Therefore:

The Polyhedron has 12 edges

pickupchik [31]3 years ago

4 0

Answer:

A polyhedron with 6 faces and 8 vertices has 12 edges.

Step-by-step explanation:

Euler Formula is given in the problem:

F + V = E + 2

We are told that the polyhedron has 6 faces (F = 6) and 8 vertices (V = 8). The only unknown variable in Euler's Formula now is E = number of edges.

Substitute the known variables into the formula and solve for E.

6 + 8 = E + 2
14 = E + 2
12 = E
E = 12

A polyhedron with 6 faces and 8 vertices has 12 edges.

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Finding Distance in the Coordinate Plane

GuDViN [60]

The distance between the two points (-1, 4) and (5, 4) is 6 units.

Step-by-step explanation:

The given Coordinate points are (-1, 4) and (5, 4).

The points are considered as (x1,y1) and (x2,y2).

Here,

x1 = -1 and x2 = 5

y1 = 4 and y2 = 4

The distance formula is given by :

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5 0

3 years ago

On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

$A = P \left(1 + \dfrac{r}{n}\right)^{nt}$

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = -1 000.00

P = 22 944.00

1 Jan 2016 A = £23 517.60

5 0

3 years ago