an object is dropped from the top of a cliff 675 meters high. its height above the ground t seconds after it is dropped is 675-4.9t^2. determine its speed 7 seconds after it is dropped, the speed if the object 7 seconds after it's dropped in m/sec

V= 68.6 m/s

Step-by-step explanation:

height above the ground t seconds after it is dropped is 675-4.9t^2

For seven seconds

Height above ground= 675-4.9(7)²

height above ground= 675-240.1

Height above ground= 434.9 m

But it's dropped from a cliff and we need to find the distance traveled from the too of the cliff

Distance= 675-434.9.

Distance= 240.1

To determine it's speed

V²= U² +2gs

V= velocity we are looking for

U= initial velocity= 0

G= acceleration due to gravity= 9.8 m/s²

s= distance covered

V²= U² +2gs

V²= 0+2(9.8)(240.1)

V²= 4705.96

V= 68.6 m/s

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.