Can 9,999,999 have the same value

The quadratic function h(t)=-16.1t^2 + 150 models a balls height, in feet, over time, in seconds, after it is dropped from a 15

katrin [286]

<h2>Hello!</h2>

The answer is: The first graphic representation.

We are given a quadratic equation, meaning that it could be two possible solutions for the exercise, however, we are talking about time, so we have to consider only the obtained positive values.

Let's make the equation equal to 0 in order to find the values of "t"

$h(t)=-16.1t^2 + 150\\0=-16.1t^2 + 150\\16.1t^2=150\\t^2=\frac{150}{16.1}=9.32\\t=+-\sqrt{9.32}=+-3.05\\t1=3.05\\t2=-305$

So, discarding the negative value, we can use the possitive value to find the correct graphic representation.

To find the correct graphic representation we must take into consideration the following:

- We must remember that the sign of the coefficient of the quadratic term (t^2) will define if the parabola opens downward or upward.

From the given quadratic (or parabola) equation we have:

$a=-1\\b=0\\c=150$

So, since the coefficient of the quadratic term is negative, the parabola opens downward.

- Since we are looking for a graphic that represents the change in height over time, we need to look for a graphic that shows only positive values for the x-axis (time)

- We are looking for a parabola which y-axis intercept is equal to 150.

Therefore, the graphic representation of the quadratic function that models a ball's height over time is the first graphic representation.

Have a nice day!

8 0

3 years ago

Read 2 more answers

You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which

stellarik [79]

To model this situation, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
$P$ is the initial deposit
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year
$t$ is the time in years

For account A:
We know for our problem that $P=2000$ and $r= \frac{2.25}{100} =0.0225$ . Since the interest is compounded monthly, it is compounded 12 times per year; therefore, $n=12$ . Lets replace those values in our formula:
$A=2000(1+ \frac{0.0225}{12} )^{12t}$

For account B:
$P=2000$ , $r= \frac{3}{100} =0.03$ , $n=12$ . Lest replace those values in our formula:
$A=2000(1+ \frac{0.03}{12} )^{12t}$

Since we want to find the time, $t$ , <span>when the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
</span> $2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000$

Now that we have our equation, we just need to solve for $t$ :
$2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000$
$(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}$
$(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}
{2}$
$ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})$
$12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})$
$t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})$
$t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}$
$17.47$

We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>

5 0

3 years ago

If f (x)=3^5-x + 6, what is the value of f (5), to the nearest tenth?

Amiraneli [1.4K]

Answer:

240

Step-by-step explanation:

3^5 - 5 + 6

243 - 5 = 238+6= 244

240?

8 0

3 years ago

Status

blsea [12.9K]

Answer:

y = -2x - 1

Step-by-step explanation:

y + 3 = -2 ( x - 1 )

y + 3 = -2x + 2

y = -2x - 1

5 0

3 years ago

Find the 50th term of 15, 9, 3, -3,...

MA_775_DIABLO [31]

Answer:-32 I THINK IM NOT 100 PERCENT POSITIVE

Step-by-step explanation:

8 0

3 years ago