I've really been stuck on these questions:
1. If ( x + 2 y )= -4 and am = 6, then 2amx + 4amy = ____ ?
2. The integers x and y are such that x > y > 0 and x 2 - y 2 = 31. If x – y = 1, what is the value of x + y ?
3. What is the value of a that will make this system of equations dependent?
: 4x=2y+10, and ax-y=5
Is it 5?
4. If (x + 1)^2 - 7 = -10, then x could be: My choices are 5, -10, 7, -7, or none of the above.
I think it is none of the above.
5. 2x^2 + bx + 3 = 0. Find the value for b that will given the above equation exactly 1 solution for x .
12?
1. start with
2amx + 4amy
= 2am(x+2y)
now both am and x+2y were given, so just sub them in
2. I am sure you meant
x^2 - y^2 = 31
factor using the difference of squares
(x+y)x-y) = 31
since x-y = 1
(x+y)(1) = 31
x+y = 31
3.
first one:
4x - 2y = 10
2nd one:
ax - y = 5
2ax - 2y = 10
to be dependent, one equation is actually the same as the other
So comparing them, 2ax = 4x
2a = 4
a = 2
4. (x+1)^2 = -3
will have no real solution, so "none of the above"
5. to have one solution , b^2 - 4ac = 0
b^2 - 4(2)(3) = 0
b^2 = 24
b = ± √24 = ± 2√6
Use the table of Standard Normal proportions to find the proportion of observations from a standard Normal distribution that have z > 1.42. (Answer should be rounded to ten-thousandths; i.e. four decimal places.)
1. To solve this problem, we'll use the given equations and isolate the variables.
Given: (x + 2y) = -4 and am = 6.
We need to find the value of 2amx + 4amy.
1. Substitute the value of am into the equation: 2amx + 4amy.
2. Replace am with 6 in the equation: 2(6)x + 4(6)y.
3. Simplify: 12x + 24y.
Therefore, the answer to 2amx + 4amy is 12x + 24y.
2. Let's solve this problem step by step.
Given: x > y > 0, x^2 - y^2 = 31, and x - y = 1.
We need to find the value of x + y.
1. Use the difference of squares formula to factor x^2 - y^2 into (x + y)(x - y).
2. Substitute the value of x - y from the second equation into the factored form: (x + y)(1) = 31.
3. Simplify to: x + y = 31.
Therefore, the value of x + y is 31.
3. We need to find the value of a that will make the system of equations dependent.
Given: 4x = 2y + 10 and ax - y = 5.
To make the system dependent, the two equations must be multiples of each other. This means that the coefficients of x and y should be in the same ratio.
1. Divide the first equation by 2 to get: 2x = y + 5.
2. Now compare the coefficient of x in the first equation with the coefficient of x in the second equation: 4 and a.
3. Similarly, compare the coefficients of y: 2 and -1.
For the system to be dependent, the ratios of coefficients should be the same. In this case, 4/2 = 2/-1.
Therefore, a = -2.
4. We need to find the value of x that satisfies the given equation: (x + 1)^2 - 7 = -10.
1. Expand the square: (x^2 + 2x + 1) - 7 = -10.
2. Simplify: x^2 + 2x - 6 = -10.
3. Rearrange the equation to get it in standard quadratic form: x^2 + 2x - 6 + 10 = 0.
4. Combine like terms: x^2 + 2x + 4 = 0.
Since this equation does not factor easily, we can use the quadratic formula to find the solutions.
5. Using the quadratic formula: x = (-b ± √(b^2 - 4ac))/2a.
In this case, a = 1, b = 2, and c = 4.
1. Substitute the values into the formula: x = (-2 ± √(2^2 - 4(1)(4)))/2(1).
2. Simplify: x = (-2 ± √(4 - 16))/2.
3. Further simplify: x = (-2 ± √(-12))/2.
4. Since the square root of a negative number is not a real number, there are no real solutions for x.
Therefore, the answer is "none of the above."
5. We need to find the value of b that will give the equation 2x^2 + bx + 3 exactly one solution for x.
To have exactly one solution for x, the quadratic equation should have a discriminant (the term inside the square root) equal to zero.
1. The discriminant is given by: b^2 - 4ac.
2. In our case, the quadratic equation is 2x^2 + bx + 3 = 0, which means a = 2, b is the unknown, and c = 3.
3. Set the discriminant equal to zero: b^2 - 4ac = 0.
4. Substitute the values: b^2 - 4(2)(3) = 0.
5. Simplify: b^2 - 24 = 0.
6. Add 24 to both sides: b^2 = 24.
7. Take the square root of both sides: b = ±√(24).
8. Simplify: b = ±2√(6).
Thus, the value for b that will give the above equation exactly one solution for x is ±2√(6).