Mathematics College

A baseball player hits a home run. A spectator observes the trajectory of the ball and expresses the height of the ball using the equation s(t)= -(t-15) + 225. Calculate the instantaneous velocity of the ball at t=4 seconds where the distance is measured in meters.

A. 44 m/sec
B. 30 m/sec
C. 26 m/sec
D. 22 m/sec

Answers

Answer 1

Answer:

The appropriate solution is Option D (22 m/sec).

Step-by-step explanation:

The given equation is:

[tex]s(t)= -(t-15)^2 + 225[/tex]

Now,

The instantaneous velocity at t = 4 will be:

= [tex][-2(t-15)(1)+0][/tex]

= [tex]-2(4-15)[/tex]

= [tex]-2\times (-11)[/tex]

= [tex]22 \ m/sec[/tex]

Thus, the above is the correct option.

Related Questions

3) A bag contains 5 yellow marbles, 2 blue marbles, 3 red
marbles. What is the probability you pull out a yelow marble?
Fraction:
Decimal:
Percent:

Answers

Answer:

fraction- 1/2

decimal- .5

percent 50%

Step-by-step explanation:

5 yellow marbles out of 10 total marbles.
Fraction: 5/10
Decimal: 0.5
Percent: 50%

Please assist, I'm really bad with word problems

Answers

Answer:

The surface area of the balloon is [tex]S(t) = \frac{100}{49} \pi t^8[/tex]

Step-by-step explanation:

Given;

S(r) = 4πr²

where;

S is the surface area of the balloon

r is the radius of the surface

The radius of the balloon at time "t" is given as;

[tex]r(t)= \frac{5}{7} t^4 \ \ \ t\geq 0[/tex]

The surface area of the balloon as a function of time "t" is calculated as;

[tex]S(t) = 4\pi (\frac{5}{7} t^4)^2\\\\S(t) = 4\pi (\frac{25}{49} t^8)\\\\S(t) = \frac{100}{49} \pi t^8[/tex]

Initially 100 milligrams of a radioactive substance was present. After 8 hours the mass had decreased by 2%. If the rate of decay is proportional to the amount of the substance present at time t, determine the half-life of the radioactive substance. (Round your answer to one decimal place.)

Answers

For initial 100 milligrams of a radioactive substance was if decreased by 2% in 8 hours, then half-life of the radioactive substance is 108.3 hours.

Rate of change is the change in one variable due to change in other.

Given the following:

Amount of a radioactive substance initially present=100mg

After 8 hours, mass decreased by 2%

Mass after 8 hours= 100-(2% of 100)

=100-(2*100/100)

=100-2

=98

The rate of decay is proportional to the amount of the substance present at time t.

Mathematically,

Let the amount of radioactive substance is A and at time t,

[tex]\dfrac d{dt}A\ \alpha\ A\\\\[/tex]

[tex]\dfrac d{dt}A= kA\\\\\\dA=kAdt[/tex]

On integrating both sides,

[tex]\int \frac {1}{kA}dA=\int dt\\\\[/tex]

[tex]\dfrac {ln A}{k}=t +ln \ c_1\\ln A =kt+k\ ln \ c_1\\[/tex]

On taking antilogarithm both sides,

[tex]e^{ln A} =e^{kt+k\ ln \ c_1\\A=e^{kt}e^{k\ ln \ c_1}\\A=e^{kt}e^{\ ln \ c_1^k}\\A=e^{kt}c_1^k \\[/tex]

Let, [tex]c_1^k=c[/tex]

[tex]A =e^{kt}c[/tex].................................1

At t=0, A=100

Substituting values in 1

[tex]100= e^0c\\c=100[/tex]

At t= 8, A=98

[tex]98=e^{k8}100\\0.98=e^{8k}\\8k=ln0.98[/tex]

k=-0.0064.

To determine the half time of the decay.

At half time A became A/2=50 mg

Substituting in 1.

[tex]50=e^{-0.0064t}100\\ln 0.5=-0.0064t\\t=\frac{ln 0.5}{-0.0064}\\t=108.3042\\t=108.3[/tex]....................................rounded to one decimal place.

Thus, The half-life of the radioactive substance is 108.3 hr.

Learn more about rate of change, here:

https://brainly.com/question/29044610

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The following stem-and-leaf plot represents the times in minutes required for 26 co-workers to commute to work. Use the data provided to find the quartiles. Commute Times in Minutes Stem Leaves 2 1 3 5 5 5 8 3 1 5 6 7 8 9 4 2 3 4 6 6 6 8 5 2 2 4 5 7 7 8 Key: 2|1

Answers

Answer:

Q1 = 30.25 minutes

Q2 = 42.5 minutes

Q3 = 40 minutes

Step-by-step explanation:

Given, In Question

The stem and leaf diagram (key 2 I 1)

Stem | Leaves

2 | 1 3 5 5 5 8

3 | 1 5 6 7 8 9

4 | 2 3 4 6 6 6 8

5 | 2 2 4 5 7 7 8

We need to find the quartiles Q1, Q2 and Q3. We know that,

Q1 = (1/4)*(n+1)th value

Q2 = (1/2)*(n+1)th value

Q3 = (3/4)*(n+1)th value

where n is the total number of co-workers 26.

So, Q1 = (1/4)*(26+1)th value

Q1 = 6.75th value

we need to count the leaves in the plot starting from the first one until we reach the 6.75th value. So, by counting, we conclude that the 6.75th value lies between the 6th and 7th value i.e. 28 and 31.

Q1 = 28 + (31-28)*0.75

= 28 + 2.25

Q1 = 30.25 minutes

Now, Q2 = (1/2) * (26+1)th value

= 13.5th value.

From the plot, we find that the 13.5th value lies in the middle of the 13th and 14th values i.e. 42 and 43. So,

Q2 = (42+43)/2

= 85/2

Q2 = 42.5 minutes

And, Q3 = (3/4)*(26+1)th value

= 20.25th value

From the plot. we find that the 20.25th value lies somewhere between the 20th and 21st value i.e. 52 and 52. So,

Q3 = 40 + (52-52)*0.25

= 40 + 0

Q3 = 40 minutes

linar system in variables
-x-3y-2z=8,
-x+y+6z=,
x-9y-2z=4

Answers

Answer:

-x+y+6z=? You did not spcify so I just made it 0

x = -22/5, y = -4/5, z = -3/5

Step-by-step explanation:

Solve the following system: