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miss Akunina [59]
3 years ago
5

A quarterback throws a football toward a receiver from a height of 6 ft. The initial vertical velocity of the ball is 14 ft/s. A

t the same time that the ball is thrown, the receiver raises his hands to a height of 8 ft and jumps up with an initial vertical velocity of 10 ft/s. Projectile motion formula: h = -16t2 + vt + h0 t = time, in seconds, since the ball was thrown h = height, in feet, above the ground { complete the system that models the heights of the ball and the receiver's hands over time.
Mathematics
2 answers:
lilavasa [31]3 years ago
6 1

Answer:

Given the projectile formula:

h = -16t^2+vt+h_0

where, t is the time in second.

As per the statement:

A quarterback throws a football toward a receiver from a height of 6 ft. The initial vertical velocity of the ball is 14 ft/s

⇒h_0 = 6 ft and v= 14 ft/s

Substitute these given values we have;

h = -16t^2+14t+6

It is also given that at the same time that the ball is thrown, the receiver raises his hands to a height of 8 ft and jumps up with an initial vertical velocity of 10 ft/s.

⇒h_0 = 8 ft and v= 10 ft/s

then;

h = -16t^2+10t+8

Therefore, the system that models the heights of the ball and the receiver's hands over time :

h = -16t^2+14t+6

h = -16t^2+10t+8

storchak [24]3 years ago
6 0

Answer:

6,10,8

Step-by-step explanation:

h = -16t2 + vt + h0

t = time, in seconds, since the ball was thrown

h = height, in feet, above the ground

h = -16t² + 14t +

h = -16t² +

t

+

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4 0
3 years ago
Please help! I keep on getting the wrong answer!
valentina_108 [34]

Answer:

CODE: 1977.98

Step-by-step explanation:

A.

(To get the closest answer, round the circumference to the nearest ten thousandth.)

C = 2(3.14)r                      Circumference formula: C = 2πr

C = 2(3.14)(3)

C = 18.84

B.

A = (3.14)r²

A = (3.14)(3)²

A = (3.14)(9)

A ≈ 28.26

C. (It's asking for the circumference.)

C = 2(3.14)r

C = 2(3.14)(58)

C ≈ 364.24

D. (It's a linear pair, which is 180 degrees.)

4x + 2x = 180

6x = 180

x = 30

m∠ABD = 4x

m∠ABD = 4(30)

m∠ABD = 120°

E. (∠GHI & ∠JHK are vertical angles, so they are congruent.)

x + 7 = 3x - 21

28 = 2x

14 = x

F. (x = 14)

m∠JHK = 3x - 21

m∠JHK = 3(14) - 21

m∠JHK = 42 - 21

m∠JHK = 21°

G. (Supplementary - two angles that add up to 180 degrees.)

180 - 84

= 96°

CODE: E(C - D) - F(G - B) - A

CODE: 14(364.24 - 120) - 21(96 - 28.26) - 18.84

CODE: 14(244.24) - 21(67.74) - 18.84

CODE: 3419.36 - 1422.54 - 18.84

CODE: 1977.98

8 0
2 years ago
Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$
Viefleur [7K]

QUESTION 1

Given that:

f(x)=2x^2+3x-9,

g(x)=5x+11,

and

h(x)=-3x^2+1

Then;

f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)

f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1

Group similar terms;

f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1

Simplify;

f(x)-g(x)+h(x)=-x^2-2x-19

QUESTION 2

Given that;

f(x)=4x-7.

g(x)=(x+1)^2

and

s(x)=f(x)+g(x)

Substitute the functions;

s(x)=4x-7+(x+1)^2

Substitute x=3

s(3)=4(3)-7+(3+1)^2

s(3)=12-7+(4)^2

s(3)=5+16

s(3)=21

QUESTION 3

Given:

f(x)=3x+2

g(x)=x^2-5x-1

f(g(x))=f(x^2-5x-1)

This implies that;

f(g(x))=3(x^2-5x-1)+2

Expand the parenthesis;

f(g(x))=3x^2-15x-3+2

f(g(x))=3x^2-15x-1

QUESTION 4

The given function is;

f(x)=3(x-6)^2+1

Let

y=3(x-6)^2+1

\Rightarrow y-1=3(x-6)^2

\Rightarrow \frac{y-1}{3}=(x-6)^2

\Rightarrow \sqrt{\frac{y-1}{3}}=x-6

\Rightarrow x=6+\sqrt{\frac{y-1}{3}}

The range is:

\frac{y-1}{3}\ge0

y-1\ge0

y\ge1

The interval notation is;

[1,+\infty)

6 0
3 years ago
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