It cant be 7 because,
X-7>3
7-7>3
0>3
And 0 is not bigger than 3
It cant be 9 because,
X-7>3
9-7>3
2>3
And 2 is not bigger than 3
So it is 11 because,
X-7>3
11-7>3
4>3
And 4 is bigger than 3
So the answer is 3) 11
Answer:
1 2/3 = m
Step-by-step explanation:
2/3 = m+3/5 -8/5
Combine terms
2/3 = m-5/5
2/3 = m -1
Add 1 to each side
2/3 +1 = m-1+1
2/3 +3/3 = m
5/3 = m
1 2/3 = m
Answer:
Step-by-step explanation:
fu**k my di*k
Answer:
36 feet.
Step-by-step explanation:
We have been given that a ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is
. We are asked to find the maximum height of the ball.
We can see that our given equation is a downward opening parabola, so its maximum height will be the vertex of the parabola.
To find the maximum height of the ball, we need to find y-coordinate of vertex of parabola.
Let us find x-coordinate of parabola using formula
.



So, the x-coordinate of the parabola is
. Now, we will substitute
in our given equation to find y-coordinate of parabola.






Therefore, the maximum height of the ball is 36 feet.
Answer:
see the explanation
Step-by-step explanation:
we know that
The circumference of a circle is equal to

where
D is the diameter
In this problem we have


substitute and solve for D



we know that
The length of the row of quarters is equal to multiply the diameter of the quarter by the number of quarters in the row
so
For 4 quarters in the row, the length is equal to --> 
For 6 quarters in the row, the length is equal to --> 
For 12 quarters in the row, the length is equal to -> 