the height of a ball dropped from a 160 foot building after t seconds is represented by h(t)=160-16t^2.how high will the ball be
1 answer:
Answer:
The height of the ball after 3 secs of dropping is 16 feet.
Step-by-step explanation:
Given:
height from which the ball is dropped = 160 foot
Time = t seconds
Function h(t)=160-16t^2.
To Find:
High will the ball be after 3 seconds = ?
Solution:
Here the time ‘t’ is already given to us as 3 secs.
We also have the relationship between the height and time given to us in the question.
So, to find the height at which the ball will be 3 secs after dropping we have to insert 3 secs in palce of ‘t’ as follows:
h(3)=160-144
h(3)=16
Therefore, the height of the ball after 3 secs of dropping is 16 feet.
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Answer:
The answer is below
Step-by-step explanation:
a) Triangle A is attached in the image below.
The base of triangle A is 3 units and its height is 3 units. The area of a triangle is given as:
Area = (1/2) × base × height
Area of triangle A = (1/2) × base × height = (1/2) × 3 × 3 = 4.5 unit²
Area of the scaled copy = 72 unit²
Ratio of area = Area of the scaled copy / Area of triangle A = 72 unit² / 4.5 unit² = 16
Hence the scaled copy area is 16 times larger than that of triangle A.
b) For the scaled copy:
Area of the scaled copy = (1/2) × base × height = 72 unit²
base × height = 144
Since the base and height are equal
base² = 144
base = 12, also height = 12
Base of scaled copy = 12 = 4 × base of triangle A
Therefore the scale factor used is 4
