So the two numbers can be represented as x and y
so x is first and y is second
y=2(x)-8
so y=2x-8
IF THEY ARE CONSECUTIVE
If they are consecutive numbers then
x+1=y
subtitute
x+1=2x-8
subtract x from both sides
1=x-8
add 8 to both sides and get
9=x
put it into the equation and get
y=2(9)-8
y=18-8
y=10
so x=9
y=10
IF X AND Y ARE CONSECUTIVE INTEGERS (1,2,3,4 not 2.3 or 1,3,5,8)
Answer:
Let's name the digit:
x- ones digit
y - tens digit
we know that x=y-2.
Now, y can be 6,7,8,9 (the number is between 60 and 100 (so depending on your understanding of "between", 0 is also possible. but then the number would have to have -2 as its ones digit, so in any case, it's not possible).
So the possibilities with x=y-2:
64
75
86
97
Out of those 64 and 86 are even, so they can't be prime.
75 has 5 in its ones number: it's divisible by 5.
so the correct answer is 97.
Step-by-step explanation:
Hope this helps!
Answer:
greater than because positive times negative equals positive
Answer:
m<1=?
m<2=4x-20..........(1)
m<4=3x-11.............(2)
Step-by-step explanation:
we have
- m<2=m<4(corresponding angle of parrelelogram are equal)
- 4x-20=3x-11
- 4x-3x=20-11
- x=9
again.we have
<1+<2=180(co-interior angle)
<1=180-<2=180-(4×9-20)=180-16=164
Answer:
- hits the ground at x = -0.732, and x = 2.732
- only the positive solution is reasonable
Step-by-step explanation:
The acorn will hit the ground where the value of x is such that y=0. We can find these values of x by solving the quadratic using any of several means.
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<h3>graphing</h3>
The attachment shows a graphing calculator solution to the equation
-3x^2 + 6x + 6 = 0
The values of x are -0.732 and 2.732. The negative value is the point where the acorn would have originated from if its parabolic path were extrapolated backward in time. Only the positive horizontal distance is a reasonable solution.
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<h3>completing the square</h3>
We can also solve the equation algebraically. One of the simplest methods is "completing the square."
-3x^2 +6x +6 = 0
x^2 -2x = 2 . . . . . . . . divide by -3 and add 2
x^2 -2x +1 = 2 +1 . . . . add 1 to complete the square
(x -1)^2 = 3 . . . . . . . . written as a square
x -1 = ±√3 . . . . . . . take the square root
x = 1 ±√3 . . . . . . . add 1; where the acorn hits the ground
The numerical values of these solutions are approximately ...
x ≈ {-0.732, 2.732}
The solutions to the equation say the acorn hits the ground at a distance of -0.732 behind Jacob, and at a distance of 2.732 in front of Jacob. The "behind" distance represents and extrapolation of the acorn's path backward in time before Jacob threw it. Only the positive solution is reasonable.