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What else can I help you with?
How would you put 5y equals 25 in standard form?
5y=25x 25x=5y -5y 25x-5y=0
What is 5y-3y?
5y-3y = 2
What is the answer for 5y equals 4735?
If: 5y = 4735 Then: y = 947
Factor this binomial 5y to the third - 125y?
-120
What are the soluition of these epautions 4x -5y -12 and -x 5y 18?
If you mean: 4x-5y = -12 and -x+5y =18 Then the solutions of the equations are: x = 2 and y = 4
Multiply y-5y 5?
y+5y
How do you factor 4t2 - 12?
4(t^2 - 3)
What is 5y plus 5y?
5y + 5y = 10y
What is y over 5 equal to y over 6 plus 9?
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
Where can you play Lego mars mission?
no site 4t2 profolio procure
Let L be the line in R3 that consists of all scalar multiples of the vector -2 -1 2T Find the reflection of the vector v equal to 7 2 7T in the line L?
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!
-4x plus 5y equals -6 -6x - y equals 5?
Multiply the first by 3 and the second by -2
How would you put 5y equals 25 in standard form?
5y=25x 25x=5y -5y 25x-5y=0
What is 52-5y?
What is 52-5y 52 = 5Y 52 = Y 5 10.4 =Y
What is 5y- 5y?
0
What expression is equivalent to -(-5y-2)?
-3
How do you solve 4 is greater than or equal to 5 y over -2 keeping in mind that only the y is over the negative?
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.
