Pls answer this question

Dvinal [7]

Answer:

Help me with this please

7 0

3 years ago

Please help! ive been stuck on this for so long and i just keep getting frustrated can someone help walkme through this?

bagirrra123 [75]

For firework launched from height 100ft with initial velocity 150ft/sec, equation made is correct

(a) equation will be $h(t) = -16t^2+150t+100$

(b) Now we have to see when it will land. At land or ground level height h will be equal to 0. So simply plug 0 in h place in equation made in part (a)

$0 = -16t^2 + 150t + 100$

Now we have to solve this quadratic. We will use quadratic formula method to solve this equation.

$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

a = -16, b = 150, c = 100.

Plugging these values in quadratic formula we get

$t = \frac{-150 \pm \sqrt{150^2-4(-16)(100)}}{2(-16)}$

$t = \frac{-150 \pm \sqrt{22500+6400}}{-32}$

$t = \frac{-150 \pm \sqrt{28900}}{-32}$

$t = \frac{-150+170}{-32} = \frac{20}{-32} = -0.625$

time cannot be negative so we will drop this answer

then $t = \frac{-150-170}{-32} = \frac{-320}{-32} = 10$

So 10 seconds is the answer for this

(c) To make table simply plug various value for t like t =0, 2, 4, 6, 8 till 10. Plug values in equation mad in part (a) and find h value for each t as shown

For t =0 seconds, $h = -16(0)^2+150(0)+100 = 100 feet$

For t =2 seconds, $h = -16(2)^2+150(2)+100 =336 feet$

For t =4 seconds, $h = -16(4)^2+150(4)+100 = 444 feet$

For t =6 seconds, $h = -16(6)^2+150(6)+100 = 424 feet$

For t =8 seconds, $h = -16(8)^2+150(8)+100 = 276 feet$

For t =10 seconds, $h = -16(10)^2+150(10)+100 = 0 feet$

(d) Axis of symmetry is given by formula

$x = \frac{-b}{2a}$

$t = \frac{-150}{2(-16)} =\frac{-150}{-32} = 4.6875$

t = 4.6875 is axis of symmetry line

(e) x-coordinate of vertex is again given by formula

$x = \frac{-b}{2a}$

so t = 4.6875

then to find y coordinate we will plug this value of t as 4.6875 in equation made in part (a)

For t =4.6875, $h = -16(4.6875)^2+150(4.6875)+100 = 451.563$

so vertex is at (4.6875, 451.563)

(f) As the firework is launched so in starting time is t=0, we cannot have time before t=0 (negative values) practically. Also we cannnot have firework going down into the ground so we cannot have h value negative physically.

6 0

3 years ago

Can someone help me!!!

Eva8 [605]

$\text {Distance} = \sqrt{(7-0)^2 + (-6+3)^2} = \sqrt{7^2+(-3)^2} = \sqrt{58} = 7.6 \text{ units}$

Answer: 7.6 units

7 0

3 years ago

ABCD is a parallelogram. If m/CDA = 63, then m/ABC = ? . The diagram is not to scale

nikdorinn [45]

it’s 63 because the sides are equal on both sides

7 0

3 years ago

You don't need to construct a Venn Diagram, it is just part of the instruction. Just determine whether or not the argument is va

icang [17]

Answer:

It is valid.

Venn Diagram is attached.

I also provided a truth table.

I also provided a couple of names this argument could be called.

Please let me know if you have any questions about this question.

Thank you kindly.

Step-by-step explanation:

If a person is a poet, then that person is a loner.

If a person is a loner, then that person is a taxi driver.

If a person is a poet, then that person is a taxi drive.

This is in the form:

p->q

q->r

Therefore, p->r.

This is an example of a hypothetical syllogism.

The first thing my mind goes to when seeing this is the transitive property which says if A is related to B and B is related to C then A is related to C.

So this syllogism also has another name and it is called the principle of transitivity of implication.

Let p=person is a poet, q=person is a loner, and r=person is a taxi driver.

I'm going to make a truth table and compare the truth values of the propositions to the conclusion.

Since we have 3 variables, we are going to need 2^3=8 rows.

p q r p->q q->r (p->q) and (q->r) conclusion=(p->r)

T T T T T T T

T T F T F F F

T F T F T F T

T F F F T F F

F F F T T T T

F F T T T T T

F T F T F F T

F T T T T T T

So we are trying to see if we ever have the preposition part true and the conclusion false.

That doesn't happen so the argument is valid.

5 0

3 years ago