Answer:
At the time of launch height of the object was 60 meters.
Step-by-step explanation:
An object was launched from a platform and its height was modeled by the function,
h(x) = -5x² + 20x + 60
Where x = time or duration after the launch
At the time of launch, x = 0
So, by putting x = 0 in this equation,
h(0) = -5×(0) + 20×(0) + 60
h(0) = 60
Therefore, at the time of launch height of the object was 60 meters.
<u>Answer</u>
59°
<u>Explanation</u>
There are 2 parallel line and one transverse.
Angles in a straight line add up to 180°
∴ 180 - (2a + 3) = 180 - 2a -3
= 177 - 2a
The angle (177 - 2a) corresponds to angle a in the diagram. The two corresponding angles are equal.
∴ 177 - 2a = a
177 = a + 2a
3a = 177
a = 177/3
= 59°
Don’t have that much money buddy
Answer:
C
Step-by-step explanation:
because the solution dors not include -3 but everything less than.
First, you must find the width of GHJK. Since the area of it is 84m², divide 84 by 7.
84÷7=12
So the width of GHJK is 12m.
Next, find the scale of GHJK to LMNP. Since the height of both are already available, you can divide 21÷7 to get 3
This means that to get the width of LMNP you must multiply the width of GHJK by 3.
12×3=36.
Now, to get the area of LMNP, multiply 21 by 36.
21×36=756
So, the area of LMNP is 756m². (Don't forget the units!>
