Asked by mysterychicken
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. 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?
Answers
Answered by
Reiny
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
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
Answered by
shenita
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.)
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