Let's make the inequality.
The sum of twice a number and 2. We're adding and multiply. X will be the number. 2x+2
Is greater than 4. We'll be using the greater than sign. >4; let's put it together.
2x+2>4; let's solve. First by subtracting two on both sides. 2x>2; let's divide by 2. x>1
So, the inequality is 2x+2>4. To solve it, it is x>1.
Answer:
Step-by-step explanation:
we have that

Using a graph tool
see the attached figure
The function represent a vertical parabola that open up
the vertex is a minimum-------> is the point (-0.5, -9.3)
The x-intercepts are the points when the y coordinate is equal to zero
The x-intercepts are the points (-3.5, 0) and (2.5, 0)
so
the function cross the negative x-axis at point
therefore, the answer is
(-4, 0) and (-3, 0)
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
The answer is C. P=h + 11
Answer:
My boy or girl everyone know that its 4
Step-by-step explanation: