Answer:
<em>x = 3</em>
Step-by-step explanation:
Step 1:
Subtract 1 from both sides
2x + 1 = 7
2x + 1 - 1 = 7 - 1
<u>2x = 6</u>
(1 - 1 will cancel itself out on the side with the variable.)
Step 2:
Divide both sides by 2
2x = 6
2x ÷ 2 = 6 ÷ 2
<u>x = 6 ÷ 2</u>
(2 divide by 2, will cancel itself out on the side with the variable.)
Step 3:
Solve for x
6 ÷ 2 = 3
x = 6 ÷ 2 = 3
<u>x = 3</u>
Answer:
Step-by-step explanation:
During the first year, ABC's stock price starts at $100 and increases by 100%. This means that the amount by which the stock increased would be
100/100 × 100 = $100
The new price of the stock would be 100 + 100 = $200
During the second year, its stock price goes down 25% from its price at the end of the first year. This means that the amount by which the stock reduced is
25/100 × 200 = 0.25 × 200 = $50
Therefore, the price of the stock, in dollars, at the end of the second year is
200 - 50 = $150
Answer:
0.12195121951
Step-by-step explanation:
3.1 / 25.42 = 0.12195121951
Hey there Ag484p1stit,
Answer:
Rent = 25% x $3,400
= $850
Left now = $3,400 - $850
= $2,550
Transportation = 30% x $2,550
= $765.
She spends $765 on transportation.
Hope this helps :D
<em>~Danielle♥</em>
Answer:
2.5 sec
Step-by-step explanation:
Height of wall = 2.5 m
initial speed of ball = 14 m/s
height from which ball is kicked = 0.4 m
we calculate the speed of the ball at the height that matches the wall first
height that matches wall = 2.5 - 0.4 = 2.1 m
using = + 2as
where a = acceleration due to gravity = -9.81 m/s^2 (negative in upwards movement)
= + 2(-9.81 x 2.1)
= 196 - 41.202
= 154.8
v = = 12.44 m/s
this is the velocity of the ball at exactly the point where the wall ends.
At the maximum height, the speed of the ball becomes zero
therefore,
u = 12.44 m/s
v = 0 m/s
a = -9.81 m/s^2
t = ?
using V = U + at
0 = 12.44 - 9.81t
-12.44 = -9.81
t = -12.44/-9.81
t = 1.27 s
the maximum height the ball reaches will be gotten with
= + 2as
a = -9.81 m/s^2
0 = + 2(-9.81s)
0 = 196 - 19.62s
s = -196/-19.62 = 9.99 m. This the maximum height reached by the ball.
height from maximum height to height of ball = 9.99 - 2.5 = 7.49 m
we calculate for the time taken for the ball to travel down this height
a = 9.81 m/s^2 (positive in downwards movement)
u = 0
s = 7.49 m
using s = ut + a
7.49 = (0 x t) + (9.81 x )
7.49 = 0 + 4.9
= 7.49/4.9 = 1.53
t = = 1.23 sec
Total time spent above wall = 1.27 s + 1.23 s = 2.5 sec