To turn a decimal into a fraction move the decimal point two spots over (Ex: 0.34=34%). To turn a fraction into a decimal get the denominator to 100 by multiplying both the top and bottom by whatever number needed (must both be multiplied by the same number) or divide to top number into the bottom number of you can’t multiply.
X+3≤ -5+2x
Add 5 to the other side. (the inverse of subtraction, since 5 is negative)
x+8≤2x
Subtract x on both sides.
8≤x <- the answer
I hope this helps!
~kaikers
So basically what you have to do is
Answer:
f(x) = -x -4 or f(x) = (-x)-4
Step-by-step explanation:
Let the graph of g be a translation of 4 units down followed by a reflection in the y-axis of the graph of f(x)=x. Write a rule for g.
Transformations can be found using this general formula: f(x) = a(bx-h)+k
For this question, we want a translation down as well as a reflection.
The two values we need to use are for k, a vertical translation, and b, a reflection over the y-axis.
Since we are translating down 4 units, k = -4
Since we are reflecting across the y-axis, b = -1
So, f(x)=(-x)-4
or
f(x)= -x -4
Answer:
D.)
Step-by-step explanation:
The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts. You can graph the function and see when the graph crosses the x-axis or solve for the x-values. I will solve it via factoring and so:

Multiply the outer coefficients, in this case 1 and 6, and 1×6=6. Now let's think about all the factors of 6 we have: 6×1 and 2×3. Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5? Actually we can say 6-1=5 and 2+3=5. Let's try both.
First let's use 6 and -1 and so:

Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.
Now let's try 2 and 3 and so:

Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:
Case 1:

Case 2:

So your zero's are when x=-2 and x=-3.
D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.
~~~Brainliest Appreciated~~~