Answer:
Arectangle=27.68cm (approximately)
Step-by-step explanation:
<em>in order to find the area of the rectangle first you need to find the measure of the length and width.</em>
since both of them aren't given ,use trigonometry to find the value of the length and width.
sin30°=opposite/hypotenuse
sin30°=opposite/8cm
opposite=sin30°*8cm. >sin30°=0.5
opposite=4cm
cos30°=adjacent/hypotenuse
cos30°=adjacent/8cm
adjacent=cos30°*8cm. cos30°=0.866
adjacent=6.92cm (approximate value)
Arectangle= bh
Arectangle=4cm*6.92cn
Arectangle=27.68cm
2x - 2y = -12
4x - 7y = -15
Multiply the first equation by -2 to make the 2x the opposite of 4x:
2x - 2y = -12 x -2 = -4x + 4y = 24
Now add this to the second equation:
-4x + 4y = 24 + 4x - 7y = -15:
-3y = 9
Divide both sides by -3
y = 9 / -3
y = -3
Now you have Y, replace y in one of the equations and solve for x:
2x - 2(-3) = -12
2x +6 = -12
Subtract 6 from both sides:
2x = -18
Divide both sides by 2:
x = -18 /2
X = -9
Check:
2(-9) - 2(-3) = -12
-18 + 6 = -12 TRUE
4(-9) - 7(-3) = -15
-36 +21 = -15 TRUE
X = -9, y = -3
(-9,-3)
So speed = distance divided by time, therefore 70 miles divided by 35 minutes = 120mph. The speedboats speed was 120mph.
Hope this helps
Answer:
y=-2
Step-by-step explanation:
because it's parallel --> a=-2
therefore we have y=-2x+b
the line goes through the point (3,-6) --> (-2).3+b=-6 --> b=0
Answer:
See below
Step-by-step explanation:
<u>First Problem</u>
The ball hits the ground when
, therefore:



and 
Since the ball is in the air before it hits the ground,
(seconds) is the more appropriate choice.
<u>Second Problem</u>
The maximum height of the ball is determined when
, therefore:




This means that the height of the ball is at its maximum after 3.34 seconds:



Thus, the answer is 54.55 (meters).
<u>Third Problem</u>
Refer to the second problem
<u>Fourth Problem</u>
<u />
<u />
<u />
<u />
<u />
<u />
<u />
Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).
<u>Fifth Problem</u>
The ball will be 24 meters off the ground when
, therefore:







Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)