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melamori03 [73]

3 years ago

9

If an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t

seconds is given by the equation h=−16 t squared +vt+s, where h is in feet. If the object is propelled from a height of 12 feet with an initial velocity of 64 feet per second, its height h is given by the equation h=-16t squared+64t+12. After how many seconds is the height 72 feet?

1 answer:

Sergio039 [100]3 years ago

5 0

From the given information we can write the equation:

-16t^2 +96t + 4 = 112

-16t^2 + 96t + 4 - 112 = 0

-16t^2 + 96t - 108 = 0

simplify, change the signs, divide by -4

4t^2 - 24t + 27 = 0

You can use the qudratic formula but this will factor to

(2t-3)(2t-9) = 0

Two solutions

t = 3/2

t = 1.5 seconds at 112 ft on the way up

and

t = 9/2

t = 4.5 seconds at 112 ft on the way back down

Graphically, (green line is 112 ft)

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son4ous [18]

Answer:

The volume is 196 cubic yards

Step-by-step explanation:

Given

$Base\ Area = 7yd * 2yd$

$Height = 14yd$

Required

The volume

The volume is given as:

$Volume = Base\ Area * Height$

Where:

$Base\ Area = 7yd * 2yd$

$Height = 14yd$

So, the equation becomes

$Volume = 7yd * 2yd * 14yd$

$Volume = 196yd^3$

6 0

3 years ago

What is the probability that you would obtain a sum of 7 or a sum of 11 on the first roll? (Rolling two dice.)

crimeas [40]

Answer:

2/9

Step-by-step explanation:

Total outcome = 36

Sum of 7 = 6

Prob of sum of 7 = 6/36

Sum of 11 = 2

Prob of sum of 11 = 2/36

Prob of sum of 7 or 11 = 6/36+2/36

Prob of sum of 7 or 11 = (6+2)/36

Prob of sum of 7 or 11 = 8/36 = 2/9

5 0

4 years ago

Estimate the quotient of 9 1/3 divided by 2 3/4

son4ous [18]

It's about 3 (3 13/33) to be exact

7 0

4 years ago

Read 2 more answers

Q3. A bag contains 7 red balls and 3 green balls. A ball is taken out replaced.

ycow [4]

Part (a)

There are 7 red out of 7+3 = 10 total

<h3>Answer: 7/10</h3>

==========================================================

Part (b)

We have 3 green out of 10 total

<h3>Answer: 3/10</h3>

==========================================================

Part (c)

3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)

So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.

<h3>Answer: 9/100</h3>

==========================================================

Part (d)

Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.

(7/10)*(7/10) = 49/100

<h3>Answer: 49/100</h3>

==========================================================

Part (e)

7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.

(7/10)*(3/10) = 21/100

<h3>Answer: 21/100</h3>

==========================================================

Part (f)

We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).

<h3>Answer: 21/100</h3>

7 0

3 years ago

Learning Task 3

ASHA 777 [7]

Answer:

1. The sides are: 38.5, 39.5, 40.5 and 41.5

2. Presently, Inzo is 15 years old and Miz is 10 years old

3. $Distance = 3682km$

4. Possible pairs are: (12, 14), (16, 18) and (20, 24)

5. $L < 38m$

Step-by-step explanation:

The question has lots of missing details. See the comment section for complete question

Solving (a):

Given

Shape: Quadrilateral

Sides: Consecutive

$Perimeter = 160$

Let one of the sides be x. The other sides will be x + 1, x + 2 and x + 3.

So, the perimeter is:

$Perimeter = x + x + 1 + x + 2 + x + 3$

This gives:

$160 = x + x + 1 + x + 2 + x + 3$

Collect Like Terms

$160 -1-2-3= x+x+x+x$

$154= 4x$

Divide both sides by 4

$38.5 = x$

$x =38.5$

<em>So, the sides are: 38.5, 39.5, 40.5 and 41.5</em>

<em></em>

Solving (b):

Given

Presently: $Inzo = 10 + Miz$

Five years ago: $Miz - 5= \frac{1}{3} * (Inzo-5)$ --- (Missing from the question)

Required: Determine their present ages

Substitute $10 + Miz$ for Inzo in the second equation

$Miz - 5 = \frac{1}{3} * (10 + Miz - 5)$

$Miz - 5 = \frac{1}{3} * (Miz + 5)$

Multiply both sides by 3

$3Miz - 15 = Miz + 5$

Collect Like Terms

$3Miz -Miz = 15 + 5$

$2Miz= 20$

Divide both sides by 2

$Miz = 10$

Substitute 10 for Miz in $Inzo = 10 + Miz$

$Inzo = 10 + 5$

$Inzo = 15$

<em>So presently, Inzo is 15 years old and Miz is 10 years old</em>

Solving (3):

Given

Express Train:

$Speed = 100kph$

$Time = t$

Local Train:

$Speed = 45kph$

$Time = t + 45$

Required: Determine the distance between A and B

Distance is calculated as:

$Distance = Speed * Time$

For the express train:

$Distance = 100 * t$

For the local train

$Distance = 45 * (45 + t)$

Distance between both stations are equal; So,

$100 * t = 45 * (45 + t)$

$100 t = 2025 + 45t$

Collect Like Terms

$100 t -45t= 2025$

$55t= 2025$

Divide by 55

$t = 36.82$

Substitute 36.82 for t in $Distance = 100 * t$

$Distance = 100 * 36.82$

$Distance = 3682km$

Solving (4):

Let the even numbers be: n and n + 2.

Sum greater than 24: $n + n + 2 > 24$

Sum less than 60: $n + n + 2 < 60$

So, we have:

$n + n + 2 > 24$

$2n + 2 > 24$

Collect Like Terms

$2n > 24-2$

$2n > 22$

$n > 11$

$n + n + 2 < 60$

$2n + 2< 60$

Collect Like Terms

$2n < 60-2$

$2n < 58$

Divide by 2

$n$

So, we have:

$n > 11$ and $n < 29$

So, possible pairs are: (12, 14), (16, 18) and (20, 24)

Solving (5):

Given

$Perimeter < 152m$

Required: Find the possible lengths

Let the length be L.

Perimeter is calculated as:

$Perimeter =4 * L$

$Perimeter =4 L$

So, we have:

$4L < 152m$

Divide through by 4

$L < 38m$

Hence, length is less than 38m (e.g. 37m)

5 0

3 years ago