5y=25x 25x=5y -5y 25x-5y=0
5y-3y = 2
If: 5y = 4735 Then: y = 947
3x/7 + 5y/14x = 3x*2x/(7*2x) + 5y/14x = (6x2 + 5y)/14x
-120
y+5y
4(t^2 - 3)
5y + 5y = 10y
no site 4t2 profolio procure
y/5 = y/6+9 Multiply all terms by 30 to eliminate the fractions 6y = 5y+ 270 Subtract 5y from both sides 6y-5y = 270 y = 270
I'm not going to lie, I'm not 100% sure if I'm going to do this correctly. It's been a while since I've done something like this, so you may want to double check my answer. Also, the letter T in your question is really throwing me off, so I'm just going to give you two answers. The first answer will treat T as a variable, the second answer will ignore it completely. Both answers, however, use the following equation for the reflection of a vector about a line:RefL(v) = 2L(v ● L)/(L ● L) - vFor my first answer, I'll use the following vectors for Land v:L = -2i - j + 2Tk, andv = 7i + 2j + 7Tk,where i, j, and k are the unit vectors in the direction of the x, y, and z axes in R3, respectively.Thus,v ● L = -14 - 2 + 14T2 = 14T2 - 16.L ● L = 4 + 1 + 4T2 = 4T2 + 5.Therefore, 2L(v ● L)/(L ● L) =2L(14T2 - 16)/(4T2 + 5) = L(28T2 - 32)/(4T2 + 5) =-8(7T2 - 8)/(4T2 + 5)i - 4(7T2 - 8)/(4T2 + 5)j + 8T(7T2 - 8)/(4T2 + 5)k.Let A = 2L(v ● L)/(L ● L).(A - v)i = [-8(7T2 - 8)/(4T2 + 5) - 7(4T2 + 5)/(4T2 + 5)]i =(-56T2 + 64 - 28T2 - 35)/(4T2 + 5)i = (-84T2 + 29)/(4T2 + 5)i.(A - v)j = [-4(7T2 - 8)/(4T2 + 5) - 2(4T2 + 5)/(4T2 + 5)]j =(-28T2 + 32 - 8T2 - 10)/(4T2 + 5)j = (-36T2 + 22)/(4T2 + 5)j.(A - v)k = [8T(7T2 - 8)/(4T2 + 5) - 7T(4T2 + 5)/(4T2 + 5)]k =(56T3 - 64T - 28T3 - 35T)/(4T2 + 5)k = (28T3 - 99T)/(4T2 + 5)k.Let b = 1/(4T2 + 5), thenRefL(v) = b[(-84T2 + 29)i + (-36T2 + 22)j + (28T3 - 99T)k]That expression for RefL(v) looks pretty ugly, so I'm going to do the problem again, this time without the variable T.L = -2i - j + 2k, andv = 7i + 2j + 7k.v ● L = -14 - 2 + 14 = -2L ● L = 4 + 1 + 4 = 9Therefore, 2L(v ● L)/(L ● L) =2L(-2/9) = L(-4/9) =(8/9)i + (4/9)j - (8/9)k.RefL(v) = 2L(v ● L)/(L ● L) - v = (8/9)i + (4/9)j - (8/9)k - 7i - 2j - 7k =-(55/9)i - (14/9)j - (71/9)k.While this expression does look much nicer, I'm not sure if it's right. So, like I recommended above, please double check my work!
Multiply the first by 3 and the second by -2
5y=25x 25x=5y -5y 25x-5y=0
What is 52-5y 52 = 5Y 52 = Y 5 10.4 =Y
0
To solve 4 ≥ 5y/-2 assuming normal algebraic rules regarding numbers and letters that are next to each other meaning multiply (and not as in a mixed number fraction): 4 ≥ 5y/-2 → 4 ≥ 5y/-2 (multiplying a fraction by a whole number → multiply numerator by the number) → 4 ≥ -5y/2 (make the denominator positive, so negate the numerator) → 8 ≥ -5y (multiply both sides by 2)* → 5y ≥ -8 (take the -5y and the 8 to the opposite sides, changing their signs) → y ≥ -8/5 (divide both sides by 5) Thus any value of y greater than or equal to -8/5 = -13/5 is a solution. * Multiplying (or dividing) both sides of an inequality by a negative number changes the direction of the inequality: greater than becomes less than and vice verse. By ensuring that the top of the negative "fraction" is the negative number and the bottom is positive, this change of inequality does not need to be remembered as a negative number is not used to multiply both sides.
you pick any # for x or y and work from there ex. x=-5 now the question is 3(-5)+5y=15 multiply 3 and -5 to get -15 add 15 to both sides and the question is now 5y=30 divide both sides by 5 and the answer is y=6