Alex, Mark & Peter share some money in the ratio 4 : 7 : 1. In total, Alex and Peter receive £60. How much does Mark get?

Molly works at an electronics store. Her last customer purchased a DVD player for $67.43 and five DVDs for $57.50, including tax

Reptile [31]

C. 66.27 is the answer

5 0

2 years ago

Please help me, i’m about to fail and i have another class on top of this !! you’ll save me.

Nataly_w [17]

((There has been a malfunction or this users account may have been deleted)))....

7 0

3 years ago

take a square of arbitary measure assuming its area is one square unit.divide it in to four equal parts and shade one of them.ag

BabaBlast [244]

Answer:

In recreational mathematics, a square array of numbers, usually positive integers, is called a magic square if the sums of the numbers in each row, each column, and both main diagonals are the same.[1][2] The order of the magic square is the number of integers along one side (n), and the constant sum is called the magic constant. If the array includes just the positive integers {\displaystyle 1,2,...,n^{2}}{\displaystyle 1,2,...,n^{2}}, the magic square is said to be normal. Some authors take magic square to mean normal magic square.[3]

The smallest (and unique up to rotation and reflection) non-trivial case of a magic square, order 3

Magic squares that include repeated entries do not fall under this definition and are referred to as trivial. Some well-known examples, including the Sagrada Família magic square and the Parker square are trivial in this sense. When all the rows and columns but not both diagonals sum to the magic constant we have semimagic squares (sometimes called orthomagic squares).

The mathematical study of magic squares typically deals with its construction, classification, and enumeration. Although completely general methods for producing all the magic squares of all orders do not exist, historically three general techniques have been discovered: by bordering method, by making composite magic squares, and by adding two preliminary squares. There are also more specific strategies like the continuous enumeration method that reproduces specific patterns. Magic squares are generally classified according to their order n as: odd if n is odd, evenly even (also referred to as "doubly even") if n is a multiple of 4, oddly even (also known as "singly even") if n is any other even number. This classification is based on different techniques required to construct odd, evenly even, and oddly even squares. Beside this, depending on further properties, magic squares are also classified as associative magic squares, pandiagonal magic squares, most-perfect magic squares, and so on. More challengingly, attempts have also been made to classify all the magic squares of a given order as transformations of a smaller set of squares. Except for n ≤ 5, the enumeration of higher order magic squares is still an open challenge. The enumeration of most-perfect magic squares of any order was only accomplished in the late 20th century.

Magic squares have a long history, dating back to at least 190 BCE in China. At various times they have acquired occult or mythical significance, and have appeared as symbols in works of art. In modern times they have been generalized a number of ways, including using extra or different constraints, multiplying instead of adding cells, using alternate shapes or more than two dimensions, and replacing numbers with shapes and addition with geometric operations.

4 0

3 years ago

Could someone please help me with this problem?<br>I keep getting the wrong answer

ahrayia [7]

When given an angle, the side opposite the angle, and another side, you have to determine how many triangles are possible or if it is not possible.

Begin with when a triangle cannot exist:
1.) measure of non-inclusive angle is less than 90° and side opposite the angle < (other side)×sin(angle measure)
2.) measure of non-inclusive angle is ≥ 90° and side opposite the angle < other side

Next determine if two triangles exist:
Only if the measure of angle is less than 90° and
(other side)×sin(non-inclusive angle measure) < opposite side < other side

Otherwise 1 triangle exist...
HERE IS WHAT YOUR PROBLEM HAS:
Non-inclusive angle measure = 22° which is < 90°
Opposite side = 13
Other side = 18.1
(other side)×sin(non-inclusive angle measure) =(18.1)×sin(22°) ≈ 6.78

So how many triangles?
6.78 < 13 < 18.1 so 2 triangles exist

Now let's find them... find angle B with law of sines
$\frac{sin22^o}{13} = \frac{sinB}{18.1}$

Put the following in your calculator: $sin^{-1}( \frac{18.1sin(22^o)}{13})$
This gives you the first angle B value and if you subtract it from 180 you get the other angle B value

To find the angle C for the first triangle 180 - (sum of angle A and first angle B)
Then use the law of sines to find side c for first triangle.

To find the angle C for the 2nd triangle 180 - (sum of angle A and 2nd angle B)
Then use the law of sines to find side c for 2nd triangle.

Sorry, I didn't do the calculations because my calculator is dead.

4 0

3 years ago

Please help me with question number 2

Mice21 [21]

Answer:

2. The area of the side walk is approximately 217 m²

3. The distance away from the sprinkler the water can spread is approximately 11 feet

4. The area of the rug is 49.6

Step-by-step explanation:

2. The dimensions of the flower bed and the sidewalks are;

The diameter of the flower bed = 20 meters

The width of the circular side walk, x = 3 meters

Therefore, the diameter of the outer edge of the side walk, D, is given as follows

D = d + 2·x (The width of the side walk is applied to both side of the circular diameter)

∴ D = 20 + 2×3 = 26

The area of the side walk = The area of the sidewalk and the side walk = The area of the flower bed

∴ The area of the side walk, A = π·D²/4 - π·d²/4

∴ A = 3.14 × 26²/4 - 3.14 × 20²/4 = 216.66

By rounding to the nearest whole number, the area of the side walk, A ≈ 217 m²

3. Given that the area formed by the circular pattern, A = 379.94 ft.², we have;

Area of a circle = π·r²

∴ Where 'r' represents how far it can spread, we have;

π·r² = 379.94

r = √(379.94 ft.²/π) ≈ 10.997211 ft.

Therefore, the distance away from the sprinkler the water can spread, r ≈ 11 feet

4. The circumference of the rug = 24.8 meters

The circumference of a circle, C = 2·π·r

Where;

r = The radius of the circle

π = 3.1

∴ For the rug of radius 'r', C = 2·π·r = 24.8

r = 24.8/(2·π) = 12.4/π = 12.4/3.1 = 4

The area = π·r²

∴ The area of the rug = 3.1 × 4² = 49.6.

8 0

3 years ago