The rocket will reach the ground after 4 seconds
Step-by-step explanation:
- Step 1: Find the time when the water rocket reaches the ground.
When the water rocket reaches the ground, the height, h will be zero.
⇒ h = 20 - 5(t - 2)²
⇒ 0 = 20 - 5(t - 2)²
⇒ 5 (t - 2)² = 20
⇒ (t - 2)² = 4
⇒ t - 2 = √4 = 2
∴ t = 2 + 2 = 4 seconds
Answer:
x=−3
Step-by-step explanation:
Move all terms that don't contain
x to the right side and solve.
hope this helps !
1.
and x+y=6
2. x^2+y^2 and x+y remind us of the formula:
substituting we get:
2xy=36-28
xy=8/2=4
3 Let's solve the system of equations i)x+y=6 and ii)xy=4
substitute y=4/x in (i):
multiply all sides by x:
complete the square:
or
or
so from x+y=6, y=6-x
case 1,
gives
case 2,
gives
Answer:
,
Answer: a) yes b) yes c) no d) no
Step-by-step explanation:
in table a there is a repeating pattern.
1 x 4 = 4
2 x 4 = 8
8 x 4 = 12
4 x 4 = 16
in table b there is a repeating pattern.
-4 x 5 = 20
-3 x 5 = 15
-2 x 5 = 10
-1 x 5 = 5
In table c there is no repeating pattern.
2 x 12 = 24
4 x 3 = 13
In table c and d what their multiplying it by is different each time than in tables a and b.
Answer:
x = sqrt(15) - 1 or x = -1 - sqrt(15)
Step-by-step explanation:
Solve for x:
x^2 + 2 x + 7 = 21
Subtract 7 from both sides:
x^2 + 2 x = 14
Add 1 to both sides:
x^2 + 2 x + 1 = 15
Write the left hand side as a square:
(x + 1)^2 = 15
Take the square root of both sides:
x + 1 = sqrt(15) or x + 1 = -sqrt(15)
Subtract 1 from both sides:
x = sqrt(15) - 1 or x + 1 = -sqrt(15)
Subtract 1 from both sides:
Answer: x = sqrt(15) - 1 or x = -1 - sqrt(15)
