<span>x^2+y^2=49
</span>
Radius 7
Center (0,0)
<span>x^2+y^2=324
</span>
Radius 18
Center (0,0)
<span>x^2+(y+2)^2=121
Radius 11
Center (0, -2)
</span><span>
(x+10)^2+(y+9)^2=8
</span>
Radius 2√2≈2.82843
Center (-10, -9)
Well a percent is always a number/100 (3%=3/100)
So what you have to do is put 66.6 over 100 and get
66.6/100 as your answer.
A sequence of transformations maps triangle ABC into triangle A’B’C
Answer:
<h2>
</h2>
Step-by-step explanation:
Divide both sides of the equation by 3
Write 7x as a difference
Factor out 3x from the expression
Factor out -4 from the expression
Factor out 2x + 5 from the expression
When the product of factors equals 0 , at least one factor is 0
Solve the equation for X
Move constant to R.H.S and change its sign
Calculate the difference
Divide both sides of the equation by 2
Calculate
Again,
Move constant to RHS and change its sign
Calculate the sum
divide both sides of the equation by 3
Calculate
Hope this helps..
Best regards!!
Answer:
0.293 s
Step-by-step explanation:
Using equations of motion,
y = 66.1 cm = 0.661 m
v = final velocity at maximum height = 0 m/s
g = - 9.8 m/s²
t = ?
u = initial takeoff velocity from the ground = ?
First of, we calculate the initial velocity
v² = u² + 2gy
0² = u² - 2(9.8)(0.661)
u² = 12.9556
u = 3.60 m/s
Then we can calculate the two time periods at which the basketball player reaches ths height that corresponds with the top 10.5 cm of his jump.
The top 10.5 cm of his journey starts from (66.1 - 10.5) = 55.6 cm = 0.556 m
y = 0.556 m
u = 3.60 m/s
g = - 9.8 m/s²
t = ?
y = ut + (1/2)gt²
0.556 = 3.6t - 4.9t²
4.9t² - 3.6t + 0.556 = 0
Solving the quadratic equation
t = 0.514 s or 0.221 s
So, the two time periods that the basketball player reaches the height that corresponds to the top 10.5 cm of his jump are
0.221 s, on his way to maximum height and
0.514 s, on his way back down (counting t = 0 s from when the basketball player leaves the ground).
Time spent in the upper 10.5 cm of the jump = 0.514 - 0.221 = 0.293 s.
