Answer:
It is (-4,3)
Step-by-step explanation:
Step-by-step explanation:
The number of tickets sold by Dan the first day is x.
Tickets sold on the 2nd day: x+3
on the 3rd day: x+6
on the 4th day: x+9
on the 5th: x+12
Since the total of tickets sold by Dan is 40:
x+(x+3)+(x+6)+(x+9)+(x+12)=40
5x+30=40
5x=10
x=2
number of tickets sold on the third day: 8
Answer:
The domain of f is {x : x ∈ R, -7 ≤ x ≤ 7} ⇒ [-7, 7]
Step-by-step explanation:
<em>The domain of a function is </em><em>all the values of x</em><em> make the function defined</em>
In the given graph
∵ The graph drawn from x = -7 to x = 7 and from y = -6 to y = 4
→ That means the values of x started from -7 and ended at x = 7
∵ The coordinates of the starting point of the graph of f are (-7, -5)
∵ The coordinates of the ending point of the graph of f are (7, 3)
∴ -7 ≤ x ≤ 7
∵ Values of x are the domain of the function
∴ The domain of the function is -7 ≤ x ≤ 7
∴ The domain of f = {x : x ∈ R, -7 ≤ x ≤ 7} ⇒ [-7, 7]
you have a quadratic equation that can be factored, like x2+5x+6=0.This can be factored into(x+2)(x+3)=0.
So the solutions are x=-2 and x=-3.
2.
<span><span>1. Try first to solve the equation by factoring. Be sure that your equation is in standard form (ax2+bx+c=0) before you start your factoring attempt. Don't waste a lot of time trying to factor your equation; if you can't get it factored in less than 60 seconds, move on to another method.
</span><span>2. Next, look at the side of the equation containing the variable. Is that side a perfect square? If it is, then you can solve the equation by taking the square root of both sides of the equation. Don't forget to include a ± sign in your equation once you have taken the square root.
3.</span>Next, if the coefficient of the squared term is 1 and the coefficient of the linear (middle) term is even, completing the square is a good method to use.
4.<span>Finally, the quadratic formula will work on any quadratic equation. However, if using the formula results in awkwardly large numbers under the radical sign, another method of solving may be a better choice.</span></span>
