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15 days ago

15

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48feet above the ground,

the function h(t)=−16t2+32t+48 models the height, h, of the ball above the ground as a function of time, t. Find the times the ball will be 48feet above the ground.

2 answers:

babunello [3.6K]15 days ago

5 0

$\bf \stackrel{height}{h(t)}=-16t^2+32t+48\implies \stackrel{48~ft}{~~\begin{matrix} 48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=-16t^2+32t~~\begin{matrix} +48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} \\\\\\ 0=-16t^2+32t\implies 16t^2-32t=0\implies 16t(t-2)=0\implies t= \begin{cases} 0\\ 2 \end{cases}$

At t = 0 seconds, the ball is released, and at t = 2 seconds, it reaches the same height again.

PIT_PIT [3.9K]15 days ago

5 0

Answer: $t_1=0\\t_2=2$

Step-by-step explanation:

We’re given the function $h(t)=-16t^2+32t+48$ that describes the ball’s height "h" above ground over time "t".

To determine when the ball is at 48 feet, substitute $h=48$ into the equation and solve for "t":

$48=-16t^2+32t+48\\0=-16t^2+32t+48-48\\0=-16t^2+32t$

After factoring, we find:

$0=-16t(t-2)\\t_1=0\\t_2=2$

You might be interested in

Find the slope of a pipe that slopes down 2/5 inch per foot.

babunello [3635]

Answer:

$m=\frac{1}{30}$

Step-by-step explanation:

To determine the slope, divide the rise by the run to calculate the slope.

Note that

1 ft = 12 in

Let

y ----> the rise

x ----> the run

m ----> the slope

$m=\frac{y}{x}$

the values are given as

$y=2/5\ in$

$x=1\ ft=12\ in$

$m=\frac{y}{x}$

substituting these values gives

$m=\frac{(2/5)}{12}$

$m=\frac{2}{60}$

Simplifying further

$m=\frac{1}{30}$

3 0

12 days ago

Read 2 more answers

Keitaro walks at a pace of 3 miles per hour and runs at a pace of 6 miles per hour. Each month, he wants to complete at least 36

tester [3916]

Answer:

Option 1 is valid, which entails 2 hours of walking and 12 hours of running.

Step-by-step explanation:

The equations provided are:

3w + 6r ≥ 36

3w + 6r ≤ 90

We'll assess which options comply with these equations.

1) 2 hours walking; 12 hours running

w = 2 and r = 12

3w + 6r ≥ 36

3(2) + 6(12) ≥ 36

6+72 ≥ 36

78 ≥ 36

3w + 6r ≤ 90

3(2) + 6(12) ≤ 90

6+72 ≤ 90

78 ≤ 90

Both equations are satisfied. Option 1 is valid.

2) 4 hours walking; 3 hours running

w = 4 and r = 3

3w + 6r ≥ 36

3(4) + 6(3) ≥ 36

12+18 ≥ 36

30 ≥ 36 (this does not hold since 30 < 36)

3w + 6r ≤ 90

3(4) + 6(3) ≤ 90

12+18 ≤ 90

30 ≤ 90

Thus, Option 2 is invalid.

3) 9 hours running; 12 hours walking

w = 9 and r = 12

3w + 6r ≥ 36

3(9) + 6(12) ≥ 36

27+72 ≥ 36

99 ≥ 36

3w + 6r ≤ 90

3(9) + 6(12) ≤ 90

27+72 ≤ 90

99 ≤ 90 (this does not hold since 99 > 90)

Option 3 is invalid.

4) 12 hours walking; 10 hours running

w = 12 and r = 10

3w + 6r ≥ 36

3(12) + 6(10) ≥ 36

36+60 ≥ 36

96 ≥ 36

3w + 6r ≤ 90

3(12) + 6(10) ≤ 90

36 + 60 ≤ 90

96 ≤ 90 (this does not hold since 96 > 90)

So, Option 4 is invalid.

3 0

6 days ago

The gas tank in felizs car is 5/6 full each time he drives to or from work he uses 1/12 of a full tank of gas which equation rep

babunello [3635]

Answer:

$\frac{1}{12}x = \frac{5}{6}\\ \\$

Step-by-step explanation:

Given:

Fuel volume = 5/6
Each trip uses fuel = 1/12
Number of trips to work = x

Since 1/12 of the tank is used for each trip and the starting volume is 5/6, the equation can be represented as:

$\frac{1}{12}x = \frac{5}{6}\\ \\$

By solving the equation, we find:

x = $\frac{5}{6}$ ÷ $\frac{1}{12}$
x = $\frac{5}{6}$ × 12
x = 10

This means Felitz can make 10 trips to/from work with a tank filled to 5/6.

3 0

7 days ago

NEED HELP ON NUMBER 13 AND 14

AnnZ [3877]

Answer:

Question 13: For age groups y=1 and y=1.3, the response time is 8 microseconds.

Question 14: The club experienced losses between 11.28 and 4.88 years.

Step-by-step explanation:

Question 13:

The equation that gives the response rate R of 8 microseconds can be expressed as

$8=y^4 +2y^3 - 4y^2 -5y +14.$

Upon graphing this, we determine the solutions to be

$y=1;$ $y=1.302;$ $y=-2;$ $y=-2.302.$

We consider only positive values of y applicable in real-life scenarios.

Thus, the response is 8 microseconds solely for the age groups y=1 and y=1.3.

Question 14:

The football club incurs losses when $p(t)$

Or

$t^3 -14t^2 +20t +120$

Graphing this inequality reveals the solutions to be

$t$ and $4.88$

As only positive values for t are relevant in practical situations, we accept the second solution.

Hence, the club faced losses during the years $4.88$

5 0

5 days ago

Initially 5 grams of salt are dissolved into 10 liters of water. Brine with concentration of salt 5 grams per liter is added at

Svet_ta [4321]

The salt enters at a rate of (5 g/L)*(3 L/min) = 15 g/min.

The salt exits at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.

Thus, the total rate of salt flow, represented by $x(t)$ in grams, is defined by the differential equation,

$x'(t)=15-\dfrac{3x(t)}{10}$

which is linear. Shift the $x$ term to the right side, then multiply both sides by $e^{3t/10}$ :

$e^{3t/10}x'+\dfrac{3e^{t/10}}{10}x=15e^{3t/10}$

$\implies\left(e^{3t/10}x\right)'=15e^{3t/10}$

Next, integrate both sides and solve for $x$ :

$e^{3t/10}x=50e^{3t/10}+C$

$\implies x(t)=50+Ce^{-3t/10}$

Initially, the tank contains 5 g of salt at time $t=0$ , so we have

$5=50+C\implies C=-45$

$\implies\boxed{x(t)=50-45e^{-3t/10}}$

The duration required for the tank to contain 20 g of salt is $t$ , such that

$20=50-45e^{-3t/10}\implies t=\dfrac{20}3\ln\dfrac32\approx2.7031\,\mathrm{min}$

3 0

8 days ago