Answer:
(0,2)
Step-by-step explanation:
2:4 means one part is 2/(2+4)=1/3 of AB and the other part is 2/3 of AB
Add 1/3 of the distance from -2 to 4. (1/3)(4+2)=2. -2+2=0 The x coordinate is 0
Subtract 1/3 of the distance from 6 to -6, (1/3)6+6)=4 6-4=2 The y coordinate is 2
The point is (0,2)
chan eil agad ach sgrùdadh a dhèanamh air google!
5x-50 = 30-15x, 20x = 80, x = 4,
x=4!
Answer: -16
Step-by-step explanation:
Let the number be y
Four times a number minus twenty-one can be written as:
(4 × y) - 21 = 4y - 21
Six times the number plus eleven can be written as:
(6 × y) + 11 = 6y + 11
Combining both equations will give:
4y - 21 = 6y + 11
4y - 6y = 11 + 21
-2y = 32
y = 32/-2
y = -16
The number is -16
<u>Prove that:</u>
![\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5Csf%5C%3A%5C%3A%5Cleft%28%5Cdfrac%7Bb%5E2-c%5E2%7D%7Ba%7D%5Cright%29%5Ccos%20A%2B%5Cleft%28%5Cdfrac%7Bc%5E2-a%5E2%7D%7Bb%7D%5Cright%29%5Ccos%20B%20%2B%5Cleft%28%5Cdfrac%7Ba%5E2-b%5E2%7D%7Bc%7D%5Cright%29%5Ccos%20C%3D0)
<u>Proof: </u>
We know that, by Law of Cosines,
<u>Taking</u><u> </u><u>LHS</u>
<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>
![\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)](https://tex.z-dn.net/?f=%5Clongmapsto%5Cleft%28%5Cdfrac%7Bb%5E2-c%5E2%7D%7Ba%7D%5Cright%29%5Cleft%28%5Cdfrac%7Bb%5E2%2Bc%5E2-a%5E2%7D%7B2bc%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7Bc%5E2-a%5E2%7D%7Bb%7D%5Cright%29%5Cleft%28%5Cdfrac%7Bc%5E2%2Ba%5E2-b%5E2%7D%7B2ca%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7Ba%5E2-b%5E2%7D%7Bc%7D%5Cright%29%5Cleft%28%5Cdfrac%7Ba%5E2%2Bb%5E2-c%5E2%7D%7B2ab%7D%5Cright%29%20)
![\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)](https://tex.z-dn.net/?f=%5Clongmapsto%5Cleft%28%5Cdfrac%7B%28b%5E4-c%5E4%29-%28a%5E2b%5E2-a%5E2c%5E2%29%7D%7B2abc%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7B%28c%5E4-a%5E4%29-%28b%5E2c%5E2-a%5E2b%5E2%29%7D%7B2abc%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7B%28a%5E4-b%5E4%29-%28a%5E2c%5E2-b%5E2c%5E2%29%7D%7B2abc%7D%5Cright%29)
![\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}](https://tex.z-dn.net/?f=%5Clongmapsto%5Cdfrac%7Bb%5E4-c%5E4-a%5E2b%5E2%2Ba%5E2c%5E2%7D%7B2abc%7D%2B%5Cdfrac%7Bc%5E4-a%5E4-b%5E2c%5E2%2Ba%5E2b%5E2%7D%7B2abc%7D%2B%5Cdfrac%7Ba%5E4-b%5E4-a%5E2c%5E2%2Bb%5E2c%5E2%7D%7B2abc%7D)
<em>On combining the fractions,</em>
![\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}](https://tex.z-dn.net/?f=%5Clongmapsto%5Cdfrac%7Bb%5E4-c%5E4-a%5E2b%5E2%2Ba%5E2c%5E2%2Bc%5E4-a%5E4-b%5E2c%5E2%2Ba%5E2b%5E2%2Ba%5E4-b%5E4-a%5E2c%5E2%2Bb%5E2c%5E2%7D%7B2abc%7D)
<em>Regrouping the terms,</em>
![\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}](https://tex.z-dn.net/?f=%5Clongmapsto%5Cdfrac%7B%28a%5E4-a%5E4%29%2B%28b%5E4-b%5E4%29%2B%28c%5E4-c%5E4%29%2B%28a%5E2b%5E2-a%5E2b%5E2%29%2B%28b%5E2c%5E2-b%5E2c%5E2%29%2B%28a%5E2c%5E2-a%5E2c%5E2%29%7D%7B2abc%7D)
![\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}](https://tex.z-dn.net/?f=%5Clongmapsto%5Cdfrac%7B%280%29%2B%280%29%2B%280%29%2B%280%29%2B%280%29%2B%280%29%7D%7B2abc%7D)
![\longmapsto\bf 0=RHS](https://tex.z-dn.net/?f=%5Clongmapsto%5Cbf%200%3DRHS)
LHS = RHS proved.
600 cubic yds of dirt needs to be removed