Hello! I would love to help!
Let's start with this part of the equation: "the sum of a number and seven"
Alright. We know that x represents an unknown number. Do you see a part of the equation that could translate to "an unknown number?"
I see "a number." So let's fill X in for "a number.
Alright. So now we have "the sum of x and 7."
Next, let's remember that sum means adding. So we just need to 7 to x
X+7
So, now instead of "the sum of a number and 7" we have x+7.
Alright. Now we just have to do the "twice." When it is asking for "twice", it is asking us to multiply our answer by two. But we need to multiply both x and 7. The best way to do that is to put our "x+7" in parenthesis and put a two outside.
2(x+7)
That's your answer! 2(x+7)
Hope this helped! Comment if you have any questions!
Answer:
Coordinates of Q 
Option D is correct option.
Step-by-step explanation:
We are given:
K is the midpoint of PQ
Coordinates of P = (-9,-4)
Coordinates of K = (-1,6)
We need to find coordinates of Q 
We will use the formula of midpoint: 
We are given midpoint K and 
the coordinates of P we need to find 
the coordinates of Q.
Now, we can write
So, we get coordinates of Q
Option D is correct option.
Answer:
8.5 Yards
Step-by-step explanation:
Pythagorean Theorem: 
a = 6
b = 6
Therefore: 
= 8.485
The path is 8.5 yards long
Answer:
I think it's C but I might be
Step-by-step explanation:
For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.
