Guide :

what is one over four squared

1/4 squared

Research, Knowledge and Information :


Foursquare - Official Site


Foursquare helps you find the perfect places to go with ... drink, shop, or visit in any city in the world. Access over 75 million short tips from local experts. Search.
Read More At : foursquare.com...

How to Square Fractions: 12 Steps (with Pictures) - wikiHow


How to Square Fractions. ... one third, squared: (1/3)^2. Thanks! Yes No. Not Helpful 3 Helpful 4. ... Bring down the squared to each part, so 17 squared over 5 squared.
Read More At : www.wikihow.com...

Foursquare - Wikipedia


Over 200: Slogan(s) Foursquare ... Foursquare is principally funded by Union Square ... If such a user was joined at a location by one of their Foursquare ...
Read More At : en.wikipedia.org...

Squares and Square Roots - Math is Fun


Squares and Square Roots. First learn about Squares, then Square Roots are easy. How to Square A Number. To square a number, just multiply it by itself ...
Read More At : www.mathsisfun.com...

Four square - Wikipedia


Four square is a ball game played among four players on a ... facing the lowest ranking square, or the ranks increase as one ... unfair control over ...
Read More At : en.wikipedia.org...

What is 1.4 squared - Answers.com


What is 1.4 squared? SAVE CANCEL. already exists. Would you like to ... conversion from one to the other is not valid without some addition information. ...
Read More At : www.answers.com...

Foursquare About Page - Food, Nightlife, Entertainment


Foursquare is a technology company that uses location intelligence to build meaningful consumer ... We're proud to be funded by Union Square Ventures, Andreessen ...
Read More At : foursquare.com...

Official Rules of Four Square | Squarefour.org


Four square is played all over the world by all different ... If the ball is touched by another object which is not one of the four players or the floor, ...
Read More At : www.squarefour.org...

Quick MathSpeak Tutorial - 2004 MathSpeak Initiative


Quick MathSpeak Tutorial. ... 5 plus StartFrac x minus one-half Over x plus one ... StartSet x Sup 1 Base comma x squared comma x cubed comma x Sup 4 Base comma ...
Read More At : www.gh-mathspeak.com...

Suggested Questions And Answer :


what is the length of each side of an octagon that fits in a 9" square

The simplest way to convert the square into an octagon is to divide each of the sides of the square into 3 segments of 3". The central segment of each side will form a 3" side of the octagon, and cutting across the corners will give you a side of length 3sqrt(2)=4.2426" approx. So the octagon is irregular with four sides equal to 3" and four sides equal to 4.2426". The lengths of the adjacent sides alternate between 3" and 4.2426". For a regular octagon the centre of the square and the centre of the octagon coincide. The central angle of the octagon measures 360/8=45, so the octagon consists of 8 isosceles triangles where the vertex angle is 45 degrees. The height of each triangle is half the side of the square=4.5". To find the length of the side of the regular octagon, we consider one isosceles triangle of height 4.5" and we split it into two back-to-back right-angled triangles. The vertex angle is bisected so that we can write (a/2)/4.5=tan(22.5), where a is the side of the octagon. So a=9tan(22.5)=3.728" approx. [tan(45)=2tan(22.5)/(1-(tan(22.5)^2); tan45=1, so 1-tan(22.5)^2=2tan(22.5). Let x=tan(22.5), then x^2+2x=1, x^2+2x+1=2; (x+1)^2=2; x+1=sqrt(2); x=tan(22.5)=sqrt(2)-1; a=9(sqrt(2)-1).] So, a regular octagon will have a side length 3.728" approx. if the square has side 9".
Read More: ...

I have 365 squares to make a blanket of 90x102.

365 factorises: 5*73. So you could make a rectangle 5 by 73 squares but the blanket is 90 by 102. The nearest square to 365 is 361, which is 19 squared. That leaves you with four squares over if you make a square of 19 by 19. You could use the 4 odd ones as an extra square next to the 10th square on each side: The size of the squares would be based on a 21 by 21 square pattern so the size would be 90/21=30/7 or smaller to fit the shorter side of the blanket. I suspect the units are inches. So that would be 4.29". To allow for a border 4.25" might be better and easier to measure. You would end up with a square side 89.25" allowing for the protuberances. The main square would have a side of 80.75", but the four odd squares would give you an extra 8.5" on each side making up to 89.25".
Read More: ...

what to type for radicals?

To radicalise a square root you break the number down to its factors and then decide whether you can pair any of the factors. For example, if the number is 64, this can be broken down to 2*2*2*2*2*2, that is 3 pairs of twos. When you take the square root, each pair reduces to one single number: so we have (2*2)(2*2)(2*2) reducing to 2*2*2. This is the number 8. But it's not always that easy. Another example, 180=2*2*3*3*5=(2*2)(3*3)5. This time we can reduce the pairs to 2*3=6, but we have a 5 left. To radicalise the square root we take the reduced pairs, 6, and multiply by the square root of 5: 6sqrt(5). That is square root of 180 in radical form. Now let's take 60=2*2*3*5=(2*2)3*5. Square root is 2sqrt(15). Example: 1350=2*3*3*3*5*5=2(3*3)3(5*5). We can reduce the pairs to 3*5 but we're left with 2*3=6; so the square root in radical form is 3*5sqrt(6)=15sqrt(6). We can't radicalise negative numbers unless we involve the imaginary number i, defined as the square root of -1. But  if we wanted to radicalise square root of -60, we would radicalise 60 then follow it with i: 2sqrt(15)i or 2isqrt(15). Not all numbers can be radicalised, but that just means we can't pair any of its factors. Other roots, like cube roots, can be radicalised, too, but instead of pairs we group in threes, so cube root of 1350=2(3*3*3)5*5. Reduce the group of three 3s to just one: 3cuberoot(50).  But 64=(2*2*2)(2*2*2), so the cube root of 64 is 2*2=4 because each of three 2s reduces to 2. For fourth roots we group in fours, and so on. Radicalisation happens in algebra, too. 25(x+1)(x-2)^2= (5*5)((x-2)*(x-2))(x+1), so square root is 5(x-2)sqrt(x+1).
Read More: ...

342-173 solve this equation in base 8 (the answer in base 10 is 169)

342 -173 _____ You can't take 3 from 2 (2 is less than 3) so you look at the 4 in the eights place. Now that's really 4 eights, so you make it 3 eights, regroup, then you change the 8 to 8 ones, then you add 'em to the 2. You get 1 2 base 8 which is 10(10) and you take away 3. That's 7. Okay? Now, instead of 4 in the eights place, you've got 3, because you added 1 (that is to say, 8) to the 2. But you can't take 7 from 3 so you look at the sixty-fours. Sixty-four.... how did sixty-four get into it, I hear you cry? Well 64 is eight squared, don't you see. Ask a silly question, you get a silly answer. From the 3, you then use 1 to make 8 ones. You add those ones to the 3 and you get 13(8) or in other words in base 10 you have 11 and you take away 7 - and 7 from 11, that's 4! Now go back to the sixty-fours, you're left with 2 and you take away 1 from 2, and that leaves... Everybody get 1? New Math. It's so simple, that only a child can do it.
Read More: ...

The sum of the ages of two brothers is 38. Four years ago the elder brothers age was square that of the younger brother. What are the ages of the two brothers?

The sum of the ages of two brothers is 38. Four years ago the age of the elder brother was square that of the younger brother. What are the ages of the two brothers? 4 years ago, each brother would have been 4 years younger. That means that the sum of their ages would have been 8 less than the sum of therir ages now. (8 = 2 * 4) So, 8 years ago the sum of their ages would gave been 30. The only two numbers, where one of them is the square of the other, and fit in this range, are 5 and 25. The ages of the brothers are: 9 yrs and 29 yrs
Read More: ...

magic square 16 boxes each box has to give a number up to16 but when added any 4 boxes equal 34

The magic square consists of 4X4 boxes. We don't yet know whether all the numbers from 1 to 16 are there. Each row adds up to 34, so since there are 4 rows the sum of the numbers must be 4*34=136. When we add the numbers from 1 to 16 we get 136, so we now know that the square consists of all the numbers between 1 and 16. We can arrange the numbers 1 to 16 into four groups of four such that within the group there are 2 pairs of numbers {x y 17-x 17-y}. These add up to 34. We need to find 8 numbers represented by A, B, ..., H, so that all the rows, columns and two diagonals add up to 34. A+B+17-A+17-B=34, C+D+17-C+17-D=34, ... (rows) A+C+17-A+17-C=34, ... (columns) Now there's a problem, because the complement of A, for example, appears in the first row and the first column, which would imply duplication and mean that we would not be able to use up all the numbers between 1 and 16. To avoid this problem we need to consider other ways of making up the sum 34 in the columns. Let's use an example. The complement of 1 is 16 anewd its accompanying pair in the row is 2+15; but if we have the sum 1+14 we need another pair that adds up to 19 so that the sum 34 is preserved. We would perhaps need 3+16. We can't use 16 and 15 because they're being used in a row, and we can't use 14, because it's being used in a column, so we would have to use 6+13 as the next available pair. So the row pairs would be 1+16 and 2+15; the column pairs 1+14 and 6+13; and the diagonal pairs 1+12 and 10+11. Note that we've used up 10 of the 16 numbers so far. This type of logic applies to every box, apart from the middle two boxes of each side of the square (these are part of a row and column only), because they appear in a row, a column and diagonal. There are only 10 equations but 16 numbers. In the following sets, in which each of the 16 boxes is represented by a letter of the alphabet between A and P, the sum of the members of each set is 34: {A B C D} {A E I M} {A F K P} {B F J N} {C G K O} {D H L P} {D G J M} {E F G H} {I J K L} {M N O P} The sums of the numbers in the following sets in rows satisfy the magic square requirement of equalling 34. This is a list of all possibilities. However, there are also 24 ways of arranging the numbers in order. We've also seen that we need 10 out of the 16 numbers to satisfy the 34 requirement for numbers that are part of a row, column and diagonal; but the remaining numbers (the central pair of numbers on each side of the square) only use 7 out of 16. The next problem is to find out how to combine the arrangements. The vertical line divides pairs of numbers that could replace the second pair of the set. So 1 16 paired with 2 15 can also be paired with 3 14, 4 13, etc. The number of alternative pairs decreases as we move down the list. 17 X 17: 1 16 2 15 | 3 14 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 2 15 3 14 | 1 16 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 3 14 4 13 | 1 16 | 2 15 | 5 12 | 6 11 | 7 10 | 8 9 4 13 5 12 | 1 16 | 2 15 | 3 14 | 6 11 | 7 10 | 8 9 5 12 6 11 | 1 16 | 2 15 | 3 14 | 4 13 | 7 10 | 8 9 6 11 7 10 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 8 9 7 10 8 9 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 6 11 16 X 18: 1 15 2 16 | 4 14 | 5 13 | 6 12 | 7 11 | 8 10 2 14 3 15 | 5 13 | 6 12 | 7 11 | 8 10 3 13 4 14 | 2 16 | 6 12 | 7 11 | 8 10 4 12 5 13 | 2 16 | 3 15 | 7 11 | 8 10 5 11 6 12 | 2 16 | 3 15 | 4 14 | 8 10 6 10 7 11 | 2 16 | 3 15 | 4 14 | 5 13 7 9 8 10 | 2 16 | 3 15 | 4 14 | 5 13 | 6 12 15 X 19: 1 14 3 16 | 4 15 | 6 13 | 7 12 | 8 11 2 13 4 15 | 3 16 | 5 14 | 7 12 | 8 11 3 12 5 14 | 4 15 | 6 13 | 8 11 4 11 6 13 | 3 16 | 5 14 | 7 12 5 10 7 12 | 3 16 | 4 15 | 6 13 | 8 11 6 9 8 11 | 3 16 | 4 15 | 5 14 | 7 12  7 8 9 10 | 3 16 | 4 15 | 5 14 | 6 13 14 X 20: 1 13 4 16 | 5 15 | 6 14 | 8 12 | 9 11 2 12 5 15 | 4 16 | 6 14 | 7 13 | 9 11 3 11 6 14 | 4 16 | 5 15 | 7 13 | 8 12 4 10 7 13 | 5 15 | 6 14 | 8 12 | 9 11 5 9 8 12 | 4 16 | 6 14 | 7 13 | 9 11 6 8 9 11 | 4 16 | 5 15 | 7 13 13 X 21: 1 12 5 16 | 6 15 | 7 14 | 8 13 | 10 11 2 11 6 15 | 5 16 | 7 14 | 8 13 | 9 12 3 10 7 14 | 5 16 | 6 15 | 8 13 | 9 12 4 9 8 13 | 5 16 | 6 15 | 7 14 | 10 11 5 8 9 12 | 6 15 | 7 14 | 10 11 6 7 10 11 | 5 16 | 8 13 | 9 12 12 X 22: 1 11 6 16 | 7 15 | 8 14 | 9 13 | 10 12 2 10 7 15 | 6 16 | 8 14 | 9 13 | 10 12 3 9 8 14 | 6 16  4 8 9 13 | 6 16 | 7 15  5 7 10 12 | 6 16 11 X 23: 1 10 7 16 | 8 15 | 9 14 2 9 8 15 | 7 16  3 8 9 14 | 7 16  10 X 24: 1 9 8 16 To give an example of how the list could be used, let's take row 3 in 17 X 17 {3 14 4 13}. If 3 is the number in the row column and diagonal, then we need to inspect the list to find another row in a different group containing 3 that doesn't duplicate any other numbers. So we move on to 15 X 19 and we find {3 12 8 11} and another row in 13 X 21 with {3 10 5 16}. So we've used the numbers 3, 4, 5, 8, 10, 11, 12, 13, 14, 16. That leaves 1, 2, 6, 7, 9, 15. In fact, one answer is: 07 12 01 14 (see 15x19) 02 13 08 11 16 03 10 05 09 06 15 04
Read More: ...

Isolate seven circles in three squares

The picture shows one solution. The squares don't have to be tilted: However, this gives the appearance of more than four squares the same size.
Read More: ...

If a square based pyramind has sides of 20 what is the the height.

Take one of the equilateral triangles and drop a perpendicular from the top to the base. The perpendicular bisects the base and forms two back-to-back right-angled triangles. The length of the perpendicular is sqrt(20^2-10^2) by Pythagoras. That's sqrt(300)=10sqrt(3). Now view the pyramid side on and drop a perpendicular from the apex where the four equilateral triangles meet on to the square base. The perpendicular meets the base at the centre of the square. Take just one of the triangular sides and view it joining the perpendicular from the apex. We have a right-angled triangle where the hypotenuse has length 10sqrt(3) and the base is half the side of the square so its length is 10. The third side of this internal triangle is the height of the pyramid=sqrt(300-10^2) where 300 is (10sqrt(3))^2. So the height is sqrt(200)=10sqrt(2)=14.14 approx.
Read More: ...

what are the properties of the quadrilateral whose vertices are the center of the squares?

The quadrilateral so formed is a rhombus (a parallelogram in which all the sides are of equal length). The interior angles of the rhombus are 90+x and 90-x where x is the angle of the parallelogram and where 90+x are the measures of opposite angles of the rhombus. The adjacent angles add up to 180, and 90+x+90-x=180. Call the parallelogram ABCD and let angle BAD be x. Because opposite side and opposite angles of a parallelogram are equal, we only need to consider two sides. Let AB=a and AD=b be two adjacent sides. Let the bisectors of these sides be points P and Q respectively. AP=PB=a/2 and AQ=QD=b/2. Outside ABCD, draw XP, length a/2, perpendicular to AB, and YQ, length b/2, perpendicular to AD. X and Y are the centres of the squares on AB and AD. One side of the new quadrilateral is defined by the line XY. So let's find its length. AX=a/sqrt(2) and AY=b/sqrt(2), because they are hypotenuses of triangles AXP and AYQ. XY=a^2/2+b^2/2-abcosXAY (cosine rule). Angle XAY=XAP+BAD+QAY=45+x+45=90+x. cosXAY=cos(90+x)=-sinx. So XY=a^2/2+b^2/2+absinx. Let's find the length of an adjacent side of the new quadrilateral. This time we have angle CDA=180-x. CD is bisected at R and WR is the perpendicular on to CD, and WR=RD=RC=a/2 because DC is parallel and equal in length to AB. Angle YDW=360-(RDW+180-x+QDY)=360-(45+180-x+45)=90+x, and cos(90+x)=-sinx. WY=DW^2+DY^2-2DWDYcosYDW=a^2/2+b^2/2+absinx. Therefore WY=XY. Since there is symmetry in the geometry the other two sides will be the same. And the opposite interior angles are equal. Therefore we have rhombus. When you draw the diagram you may find that XY is exterior to the parallelogram, and WY is interior. Nevertheless, since sin(180-x)=sinx, the lengths of the sides will still be equal. There are various other geometries that may apply, but you will find that you always get a combination of two angles of 45 degrees and the angle of the parallelogram (skew), but sinx will always be the result (for example, cos(90-x)=sinx, cos(45-x+45)=sin(x), cos(180-(90+x))=sinx, etc.).
Read More: ...

cost of white washing on its four walls

tu get area av 4 walls av a room: leng=8, wide=6, hite=4 total leng=2*(8+6)=2*14=28 area=length*hite=28*4=112 (square meters)
Read More: ...

Tips for a great answer:

- Provide details, support with references or personal experience .
- If you need clarification, ask it in the comment box .
- It's 100% free, no registration required.
next Question || Previos Question
  • Start your question with What, Why, How, When, etc. and end with a "?"
  • Be clear and specific
  • Use proper spelling and grammar
all rights reserved to the respective owners || www.math-problems-solved.com || Terms of Use || Contact || Privacy Policy
Load time: 0.0707 seconds