If the last 2 digits of a number is divisble by 4, then the number is divisble by 4
see what combos are with ending with 2 digits divisible by 4
24
64
52
72
see their combos for the first 2 numbers
24 three other numbers to pick from so 3*2*1=6
64 three other numbers to pick from so 3*2*1=6
52 three other numbers to pick from so 3*2*1=6
72 three other numbers to pick from so 3*2*1=6
6*4=24
answer is 24 ways
Answer
Dilations is a transformation that produces an image that is the same shape as A description of a dilation includes the scale factor (or ratio) and the center of the Most dilations in the coordinate plane use the origin, (0,0), as the center of the. the center of the dilation at point A to the other points B, C and D. The dilation
Given that the last term is -3x^4, the polynomila is ordered in descending order of the exponent of x. Then, the first term is that where y is with the highest exponent, that is y^4
Simplify the terms with y^4: -2y^4 + 6y^4 = 4y^4
Then the first term is 4y^4
The base case is the claim that

which reduces to

which is true.
Assume that the inequality holds for <em>n</em> = <em>k </em>; that

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

By the induction hypothesis,

Now compare this to the upper bound we seek:

because

in turn because

Answer:
x > -9
Step-by-step explanation:
The solution to the inequality is obtained as follows:
-3x -7 < 20
-3x - 7 + 7 < 20 + 7
-3x < 27
-3x/(-3) > 27/(-3)
x > -9
Notice that when you divide by a negative number, the inequality sign changes.
In the picture attached, the graph of the solution s shown. Notice that -9 is not included in the solution.