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Which two numbers have the mean of 10 and the range of 8?
6 and 14 Solution x - y= 8, so x =8 + y, (x + y)/2 = 10, x + y = 20, Substituting x = 8 + y, we have: 2 y + 8 = 20, or 2y = 12, y = 6, x = 6+8= 14
The sum of two number is 20 and their difference 4 Find the numbers?
x+y=20 x-y=4 2x=24 x=12 y=8
What two real numbers have a sum of 8 and a product of 2?
x + y = 8 x * y = 2 First let's solve the first equation for x in terms of y: x + y = 8 x = 8 - y Now let's plug that into our second equation and simplify it: x * y = 2 (8 - y) * y = 2 8y - y2 = 2 -y2 + 8y - 2 = 0 Now the always fun quadratic equation to find y: (Since I don't know how the system will respond to a "plus or minus" sign, I'll use +/- instead) -b +/- sqrt(b2 - 4*a*c) ---------------------------- 2*a ...and plugging in our values (a = -1, b = 8, c = -2) -8 +/- sqrt(82 - 4(-1)(-2)) ------------------------------- 2(-1) -8 +/- sqrt(64 - 8) ---------------------- -2 -8 +/- sqrt(56) ------------------ -2 We'll use a calculator to figure out that sqrt(56) = 2*sqrt(14) -8 +/- 2*sqrt(14) ---------------------- -2 Please excuse the formatting... -8/-2 +/- 2*sqrt(14)/(-2) 4 +/- (-sqrt(14)) 4 -/+ sqrt(14) Note the shift in the +/- to -/+ when we multiply out that -1 y = 4 -/+ sqrt(14) Now that we know y, let's go back to our very first equation and plug in the value. x = 8 - y x = 8 - (4-/+sqrt(14)) Now we have two possible values of y, which change the possible value of x: When y = 4 - sqrt(14), then x = 4 + sqrt(14) When y = 4 + sqrt(14), then x = 4 - sqrt(14) In conclusion, the two real numbers which have a sum of eight and a product of two are: x = 4 - sqrt(14), y = 4 + sqrt(14) x = 4 + sqrt(14), y = 4 - sqrt(14)
What 2 number multiply to -120 and add up to -14?
This is algebra. Let's take 2 numbers, x and y. What we want is xy = -120 and x+y = -14. So we have two equations. Let's solve for one of the numbers and move that over to the other equation. x+y=-14 x=-14-y Let us now plug that x into the other equation. (-14-y)y=-120 We now distribute the y to both -14 and -y. -14y-y^2=-120 moving everything to one side we get 0=y^2+14y-120 now it's just a matter of solving for y, which can be found using the quadratic equation (-b+-(b^2-4ac)^(1/2))/(2a), where a=1, b=14, and c=-120. Let's try plus first (-14+(14^2-(4)(-120))^(1/2))/2 (-14+(196+480)^(1/2))/2 (-14+26)/2 6 so y=6. Let's plug that into one of the equations we had, x+y=-14. x+6=-14 x=-20 To see if this works, let's try y=6 and x=-20 on both equations -20(6)=-120 -20+6=-14 They work! By the way, if we had tried minus in the quadratic equation, we would've gotten -20 for y and x as 6.
What. A proportional relationship exists between x and y. When x=8 the value of y is 14. A) Write an equation for y in terms of x.?
Please dont mind the ”what”
Which two numbers have the mean of 10 and the range of 8?
6 and 14 Solution x - y= 8, so x =8 + y, (x + y)/2 = 10, x + y = 20, Substituting x = 8 + y, we have: 2 y + 8 = 20, or 2y = 12, y = 6, x = 6+8= 14
The sum of two number is 20 and their difference 4 Find the numbers?
x+y=20 x-y=4 2x=24 x=12 y=8
The difference between two numbers is 20 and their sum is 48 Find the numbers?
x=number 1, y=number 2 {x-y=20} {x+y=48} x=y+20 y+20+y=48 2y=28 [y=14] x=14+20 [x=34]
Where do the lines 5x 3y 20 and 2x - y -14 intersect?
I presume that you mean 5x + 3y + 20 = 0 and 2x - y - 14 = 0. First put the formulae in the form y = mx + c. 3y = -5x -20 y = -5/3 x -20/3 and y = -2x + 14 Substitute the second formula into the first -2x - 14 = -5/3 x -20/3 -6x - 42 = -5x - 20 -42 = x - 20 x = -22 Substitute into equation y = -2x + 14 y = -2 x -22 + 14 y = 44 + 14 y = 58 Therefore the two lines meet at point (-22, 58).
If 4 multiplied by y eqeals x and x added by y eqeals 10 what is the value of x and y?
14
-x+ 4 y>-8 x+y8 y?
-x+ 4y>-8, x+y≤3 y≤3-x -x+4(3-x)>-8 -x+12-4x>-8 12-5x>-8 -5x>-20 x>4 y≤3-(4) y≤-1
What is the answer to y equals 3x-1?
4
Solve using elimination x plus y equals 8 and x minus y equals 6?
X + y = 8 x - y = 6 add equations to eliminate y 2x = 14 x = 7 substitute 7 + y = 8 y = 1
What is the value of y if 3y plus 8 equals 5y-20?
If: 3y+8 = 5y-20 Then: 3y-5y = -20-8 So: -2y = -28 And: y = -28/-2 => y = 14
What two real numbers have a sum of 8 and a product of 2?
x + y = 8 x * y = 2 First let's solve the first equation for x in terms of y: x + y = 8 x = 8 - y Now let's plug that into our second equation and simplify it: x * y = 2 (8 - y) * y = 2 8y - y2 = 2 -y2 + 8y - 2 = 0 Now the always fun quadratic equation to find y: (Since I don't know how the system will respond to a "plus or minus" sign, I'll use +/- instead) -b +/- sqrt(b2 - 4*a*c) ---------------------------- 2*a ...and plugging in our values (a = -1, b = 8, c = -2) -8 +/- sqrt(82 - 4(-1)(-2)) ------------------------------- 2(-1) -8 +/- sqrt(64 - 8) ---------------------- -2 -8 +/- sqrt(56) ------------------ -2 We'll use a calculator to figure out that sqrt(56) = 2*sqrt(14) -8 +/- 2*sqrt(14) ---------------------- -2 Please excuse the formatting... -8/-2 +/- 2*sqrt(14)/(-2) 4 +/- (-sqrt(14)) 4 -/+ sqrt(14) Note the shift in the +/- to -/+ when we multiply out that -1 y = 4 -/+ sqrt(14) Now that we know y, let's go back to our very first equation and plug in the value. x = 8 - y x = 8 - (4-/+sqrt(14)) Now we have two possible values of y, which change the possible value of x: When y = 4 - sqrt(14), then x = 4 + sqrt(14) When y = 4 + sqrt(14), then x = 4 - sqrt(14) In conclusion, the two real numbers which have a sum of eight and a product of two are: x = 4 - sqrt(14), y = 4 + sqrt(14) x = 4 + sqrt(14), y = 4 - sqrt(14)
What numbers can you multiply to get 8 and subtract to get -6?
14
What 2 number multiply to -120 and add up to -14?
This is algebra. Let's take 2 numbers, x and y. What we want is xy = -120 and x+y = -14. So we have two equations. Let's solve for one of the numbers and move that over to the other equation. x+y=-14 x=-14-y Let us now plug that x into the other equation. (-14-y)y=-120 We now distribute the y to both -14 and -y. -14y-y^2=-120 moving everything to one side we get 0=y^2+14y-120 now it's just a matter of solving for y, which can be found using the quadratic equation (-b+-(b^2-4ac)^(1/2))/(2a), where a=1, b=14, and c=-120. Let's try plus first (-14+(14^2-(4)(-120))^(1/2))/2 (-14+(196+480)^(1/2))/2 (-14+26)/2 6 so y=6. Let's plug that into one of the equations we had, x+y=-14. x+6=-14 x=-20 To see if this works, let's try y=6 and x=-20 on both equations -20(6)=-120 -20+6=-14 They work! By the way, if we had tried minus in the quadratic equation, we would've gotten -20 for y and x as 6.
