Hello!
The answer should be the second option B.
Acceleration is a rate of change in velocity. It involves a change in speed and direction. Change of velocity over a period of time.
Acceleration formula
↓↓↓↓↓↓↓
A=Δv/t
Hope this helps!
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-Charlie
Answer:
270
Step-by-step explanation:
becasue 100 percent of 90 would be 90

so multiply

ok first you need to add e and 4 and get 95 out of haha
Answer:
-b/2a is the equation to find the x coordinate of the vertex point and it's y, when x is plugged into the quadratic equation.
Step-by-step explanation:
Answer:
x = 2
Step-by-step explanation:
These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.
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<h3>Squaring</h3>
The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.
x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true
x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution
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<h3>Substitution</h3>
Another way to solve this is using substitution for one of the radicals. We choose ...

Solutions to this equation are ...
u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution
The value of x is ...
x = u² -2 = 2² -2
x = 2 . . . . the solution to the equation
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<em>Additional comment</em>
Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.