0
Add your answer:
Earn +20 pts
the ratio of A to B is 2:3 and the ratio of B to C is 4:5. what is the ratio of A to C?
This deals with ratios and proportions. ⊱ ────── ✯ ────── ⊰ A : B = 2 : 3 B : C = 4 : 5. Now, to find A : B : C, we need to make the value of B equal in A : B ratio and B : C ratio. Here, Value of B in A : B ratio is 3; and B : C ratio is 4. LCM of 3 and 4 is 12. Therefore, we multiply 4 to the first ratio and 3 to the second ratio. A : B = 2 × 4 : 3 × 4 A : B = 8 : 12 Also, B : C = 4 × 3 : 5 × 3 B : C = 12 : 15 Now, we can combine A : B and B : C. A : B : C = 8 : 12 : 15.
Points a b c and d are coplanar b c and d are collinear not a how many lines are determined by a b c and d?
5 its 4
What numbers is divisible by both 9 and 4-A 24815 B 18324 C 9140 D 9126?
B
A is 2 B is 4 what is c?
"6" It all depends on the relationship between A B and C. EG if a+b=c the c=6, if axb=c then c=8. you can throw in allsorts of relationships C/A = B So more information is required in the question
What is the sun of four integers whose average is 15?
I assume you meant "sum". The answer is 60. 60=4 (integers) * 15 (average value of the integers). average of 4 integers = 15 --> (a + b + c + d)/4 = 15 --> (a + b + c + d) = 15 * 4 --> (a + b + c + d) = 60
the ratio of A to B is 2:3 and the ratio of B to C is 4:5. what is the ratio of A to C?
This deals with ratios and proportions. ⊱ ────── ✯ ────── ⊰ A : B = 2 : 3 B : C = 4 : 5. Now, to find A : B : C, we need to make the value of B equal in A : B ratio and B : C ratio. Here, Value of B in A : B ratio is 3; and B : C ratio is 4. LCM of 3 and 4 is 12. Therefore, we multiply 4 to the first ratio and 3 to the second ratio. A : B = 2 × 4 : 3 × 4 A : B = 8 : 12 Also, B : C = 4 × 3 : 5 × 3 B : C = 12 : 15 Now, we can combine A : B and B : C. A : B : C = 8 : 12 : 15.
If you have a 10 measuring cup 4 measuring cup and a 3 measuring cup how many pours does it take to get the 4 to 4 the 3 to 1 and the 10 to half?
Assuming the 10 = Cup A, 4 = Cup B and 3 = Cup C 1) Fill Cup C (A=0, B=0, C=3) 2) Pour Cup C into Cup A (A=3, B=0, C=0) 3) Fill Cup B (A=3, B=4, C=0) 4) Fill Cup C from Cup A (A=3, B=1, C=3) 5) Pour the remainder of Cup B into Cup A (A=4, B=0, C=3) 6) Empty Cup C (A=4, B=0, C=0) 7) Fill Cup B (A=4, B=4, C=0) 8) Fill Cup C from Cup A (A=4, B=1, C=3) 9) Pour the remainder of Cup B into Cup A (A=5, B=0, C=3) 10) Empty Cup C (A=5, B=0, C=0) 11) Fill Cup B (A=5, B=4, C=0) 12) Fill Cup C from Cup A (A=5, B=1, C=3) 13) Empty Cup C (A=5, B=1, C=0) 13) Pour the remainder of Cup B into Cup C (A=5, B=0, C=1) 14) Fill Cup B (A=5, B=4, C=1) so assuming you count the filling of cups as pours your answer is 14
Maria keeps her four stuffed bears lined up on a shelf over her bed how many arrangements of the bears are possible?
The answer is 4! (4 factorial), the same as 4x3x2x1, which equals 24 combinations. The answer is 24 and this is how: A b c d A b d c A c d b A c b d A d c b A d b c B c d a B c a d B d a c B d c a B a c d B a d c C d a b C d b a C a b d C a d b C b d a C b a d D a b c D a c b D b c a D b a c D c a b D c b a
What is proof of Heron's Formula?
This is a proof that uses the cosine rule and Pythagoras' theorem. As on any triangle with c being the opposite side of θ and a and b are the other sides: c^2=a^2+b^2-2abcosθ We can rearrange this for θ: θ=arccos[(a^2+b^2-c^2)/(2ab)] On a right-angle triangle cosθ=a/h. We can therefore construct a right-angle triangle with θ being one of the angles, the adjacent side being a^2+b^2-c^2 and the hypotenuse being 2ab. As the formula for the area of a triangle is also absinθ/2, when a and b being two sides and θ the angle between them, the opposite side of θ on the right-angle triangle we have constructed is 4A, with A being the area of the original triangle, as it is 2absinθ. Therefore, according to Pythagoras' theorem: (2ab)^2=(a^2+b^2-c^2)^2+(4A)^2 4a^2*b^2=(a^2+b^2-c^2)^2+16A^2 16A^2=4a^2*b^2-(a^2+b^2-c^2)^2 This is where it will start to get messy: 16A^2=4a^2*b^2-(a^2+b^2-c^2)(a^2+b^2-c^2) =4a^2*b^2-(a^4+a^2*b^2-a^2*c^2+a^2*b^2+b^4-b^2*c^2- a^2*c^2-b^2*c^2+c^4) =4a^2*b^2-(a^4+2a^2*b^2-2a^2*c^2+b^4-2b^2*c^2+c^4) =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.1) We will now see: (a+b+c)(-a+b+c)(a-b+c)(a+b-c) =(-a^2+ab+ac-ab+b^2+bc-ac+bc+c^2)(a^2+ab-ac-ab-b^2+bc+ac+bc-c^2) =(-a^2+b^2+2bc+c^2)(a^2-b^2+2bc-c^2) =-a^4+a^2*b^2-2a^2*bc+a^2*c^2+a^2*b^2-b^4+2b^3*c-b^2*c^2+2a^2*bc-2b^3*c+(2bc)^2-2bc^3+a^2*c^2-b^2*c^2+2bc^3-c^4 =-a^4+2a^2*b^2+2a^2*c^2-b^4+(2bc)^2-c^4-2b^2*c^2 =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.2) And now that we know that Eq.1=Eq.2, we can make Eq.1=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) Therefore: 16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 =[(a+b+c)/2][(-a+b+c)/2][(a-b+c)/2][(a+b-c)/2] And so if we let s=(a+b+c)/2 A^2=s(s-a)(s-b)(s-c)
If A is less than B and B plus C equals 10 and none of them equal zero then which of the following must be true?
Well, isn't that just a happy little math problem! If A is less than B and B plus C equals 10, then it must be true that A plus C is less than 10. Just remember, in the world of numbers, everything adds up beautifully in the end.
Notes to play you got to feeling?
here's the notes for i gotta feeling on the recorder by the black eyed peas: d' d' d' d' c' c' bb c'c'c'c' c'c'c'c' b b b b b b b b a ( 4 beats) g ( 4 beats) a ( 4 beats) g ( 4 beats) c' a ( 4 beats) g ( 4 beats) a ( 4 beats) g ( 4 beats) E A G C' B..........E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' B E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' B E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' B E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' B A...G. Thats the begining thnks ;)
How do you play i got to feeling on recorded?
Here's the notes for i gotta feeling on the recorder ( can also be played on the keyboard with the same notes) by the black eyed peas:d' d' d' d' c' c' bb c'c'c'c' c'c'c'c' b b b b b b b b c'c'c'c' c'c'c'c'a ( 4 beats) g ( 4 beats)a ( 4 beats) g ( 4 beats) c'a ( 4 beats) g ( 4 beats)a ( 4 beats) g ( 4 beats)E A G C' B..........E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' BE E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' BE E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' BE E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' BA...G. Thats the begining thnks ;)
What are the notes to believe by polar express on viola?
PLEASE NOTE ~ |= MEASURE SEPARATION ALL OF THE Ds ARE HIGH D AND OPEN D WILL BE WRITTEN IN ITALICS ( D )4/4 B B B B B B| B D G A B| C# C# C# C# C# B B|B A A B D|B B B B B B| B D G A B| C# C# C# C# C# B B B| D D C# A G| D B A G D | D B A G E | E C# B A F | D D C# A B| D B A G D | D B A G E | E C# B A D D D D | E-(high) D C# A G D| B B B B B B| B D G A B| C# C# C# C# C# B B|B A A B D| B B B B B B| B D G A B| C# C# C# C# C# B B B| D D C# A G|
How do you solve a equals b minus 4 over c for b?
Well. Multiply both sides by 'c'. After that you should have something like ac=b-4 then you add 4, to get this result 4+ac=b.
Associative addition math problem?
a+(b+c)=b+(a+c) 2+(7+4)=7+(2+4)
Pennfoster calculus exam 4?
1.A 2.d 3.b 4.a 5.c 6.a 7.c 8.c 9.b 10.c 11.b 12.b 13.a 14.a 15.a
Ax2 plus bx plus c equals 0?
x=(-b+sqrt(b^2-4*a*c))/(2a) and (-b-sqrt(b^2-4*a*c))/(2a) (This is called the quadratic formula)
