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do these ratios form a proportion? 4:2 = 20:6

I need to know  if each pair of ration form a proportion?

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Proportions Flashcards | Quizlet

Do these ratios form a proportion? 4/2 and 20/6. 4(6) = 24 and 2(20) = 40 so NO. Do these ratios form a proportion? 3/2 and 18/8. 3(8) = 24 and 2(18) = 36 so NO.
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Ratios and Proportions - Proportions - In Depth

... 20 and 5 are the extremes, and 25 and 4 are the ... we know that these ratios are equal and that this is a ... Ratios and Proportions: Email this page to ...
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Comparing Ratios to Form a Proportion - Math Forum

Comparing Ratios to Form a Proportion ... the following four ratios form a proportion? 3 to 4 4 to 6 8 ... ratios 3 to 15, 10 to 40, 5 to 60, and 20 to 80 I ...
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Proportions - Basic mathematics

Do they form a proportion? ... Do they form a proportion? They will form a proportion if the ratios are equal, so first put an equal sign between them
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Ratio Calculator

... ratio proportions and ratio formulas. Calculator solves ratios for the missing value or compares 2 ratios and evaluates as true or false.
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Ratio Proportions - Math is Fun - Maths Resources

Working With Proportions. NOW, how do we use ... we know into this form: ... values by the same amount and still have the same ratio. 10:20:60 is the same as 1:2:6.
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IXL - Do the ratios form a proportion? (7th grade math practice)

Fun math practice! Improve your skills with free problems in 'Do the ratios form a proportion?' and thousands of other practice lessons. Sign in Remember.
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Solving Simple Proportions (page 4 of 7) - Purplemath | Home

Solving Simple Proportions (page 4 of 7) Sections: Ratios, ... First, I convert the colon-based odds-notation ratios to fractional form:. Then I ...
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Suggested Questions And Answer :

question related to sequence and series with relation with roots of equation

a(x-x1)(x-x2)=ax^2-a(x1+x2)x+ax1x2=ax^2+bx+c=0. b=-a(x1+x2), c=ax1x2 and b^2-4ac=a^2(x1+x2)^2-4a^2x1x2=a^2(x1-x2)^2. Similarly, q=-p(x3+x4), c=px3x4 and p^2-4pr=p^2(x3-x4)^2, so (b^2-4ac)/(q^2-4pr)=(a^2(x1-x2)^2)/(p^2(x3-x4)^2) or ((a(x1-x2))/(p(x3-x4)))^2, where x3 and x4 are roots of the second polynomial. If x3 and x4 are replaced by x^3 and x^4 then the division becomes ((a(x1-x2))/(p(x^3-x^4)))^2. (i) If AP is x1, x2, 1/x3, 1/x4, then if d is the common difference, x2=x1+d so x1-x2=-d or (x1-x2)^2=d^2. If 1/x4-1x3=d, then x3-x4=dx3x4, x4(1+dx3)=x3, x4=x3/(1+dx3). Also, if 1/x3-x2=d, then 1-x2x3=dx3 and x3(d+x2)=1, x3=1/(d+x2). Let A=x1 then x2=A+d, 1/x3=A+2d, 1/x4=A+3d. x3=1/(A+2d), x4=1/(A+3d). x3-x4=d/((A+3d)(A+2d)). The division becomes: (a(A+2d)(A+3d)/p)^2=(a(x1+2d)(x1+3d)/p)^2.  (ii) If a, b, c, form a GP, then b=ar, c=ar^2, where r=common ratio. b^2-4ac=a^2r^2-4a^2r^2=-3a^2r^2. Since square root of b^2-4ac must be greater than or equal to zero for real roots of the quadratic to exist, if a, b and c are consecutive terms in a GP, then there can be no real roots. The imaginary part of the root is given by +arisqrt(3)/2a=+risqrt(3)/2 where i is the imaginary square root of -1, and x1 and x2 are both complex numbers. If x1, x2, x3, x4 form a GP with common ratio R, x1=A, x2=AR, x3=AR^2, x4=AR^3. q=-p(x3+x4)=-pAR^2(1+R) and r=px3x4=pA^2R^5. p, q, r form a series p, -pAR^2(1+R), pA^2R^5, which is not an obvious GP, because there is no visible common ratio. However, if we call the supposed common ration X, then R must be such that pX=-pAR^2(1+R) and pX^2=pA^2R^5, making X=-pA^2R^5/pAR^2(1+R)=-AR^3/(1+R) and X=-AR^2(1+R). So R^3/(1+R)=R^2(1+R); R=(1+R)^2; R^2+R+1=0, which has no real roots. The solution is a complex number: (-1+isqrt(3))/2 or (-1-isqrt(3))/2. From this, X=-A(1-3+2isqrt(3))/2isqrt(3)=x1(1+isqrt(3))/isqrt(3)=-x1(isqrt(3)/3+1). This common ratio applies only to p, q and r.  
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72:20 , 5:18 does this set of ratio form a proportion ?

72:20 is the same as 18:5 but 18:5 is not the same as 5:18 when it's applied to two quantities. The order of the numbers is important. For example, if we say the ratio of boys to girls is 5:18 (implying 23 boys and girls, or some multiple of 23, like 4*23=92), we can also say that the ratio of girls to boys is 18:5, but we can't say that the ratio of girls to boys is 5:18. We can express the ratio as a proportion: 18/23 and 5/23 imply the ratio 18:5. In the example, if there are 92 girls and boys, 72 would be girls and 20 boys. The ratio is still 18:5, 18 girls to 5 boys.
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find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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Do these ratios form a proportion? 210/30 and 343/49

?????????????? "rasheos & proporshuns" ?????????? em thangs be FRAKSHUNS
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Write = if the ratios form a proportion;if they do not form a proportion,€

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Two equal and adjacent angles forming linear pair.

?????????????? "2 angels form linear pare" ???????????????
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Two alloys contain tin and iron in the ratio of 1:2 and 2:3. If the two alloys are mixed in the proportion of 3:4 respectively (by weight). find the ratio of tin and iron in the newly formed alloy.

Let the weights of tin and iron be S and F respectively. The first alloy consists of S/3+2F/3, the second 2S/5+3F/5. The mixture contains 3(S/3+2F/3)/7+4(2S/5+3F/5)/7=(S+2F+1.6S+2.4F)/7=(2.6S+4.4F)/7=2(1.3S+2.2F)/7. Ratios of tin and iron in the mixture are 1.3:2.2 or 13:22.
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Does 1:5, 5:15 form a proportion

1:5 can be written 3:15. (5:15 is the same as 1:3 or the fraction 1/3.) So the ratio of 1:5 to 5:15 is 3:15 to 5:15 or 3:5. The ratio can be expressed as a fraction or proportion 3/5. 1/5 is 3/5 of 1/3.
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do these ratios form a proportion? 4:2 = 20:6

no...4/2=2 20/6=10/3=3.3333333333...
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ratios, proportions, and geometric means x/3=4x/x+3

x/3 = 4x/(x + 3) cross multiply: x(x + 3) = 12x distribute: x^2 + 3x = 12x subtract 12x from both sides: x^2 - 9x = 0 factor: x(x - 9) = 0 x = 0, 9
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