The question is : -8(5x-3)+2x

Jerald jumped from a bungee tower. If the equation that models his height, in feet, is h = –16t2 + 729, where t is the time in s

vekshin1

104 = -16T2 + 729

16t^2 = 729 - 104 = 625

4t = +/-25

t = +/-6.25

Between 6.25 and 13.5 seconds after he jumps

7 0

3 years ago

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Plant A: A graph has time (weeks) on the x-axis, and height (inches) on the y-axis. A line goes through points (0, 3), (1, 4.8),

butalik [34]

Answer:

The correct option is;

No, The greater rate of change of Plant A will result in it being 0.9 inches taller in 6 weeks

Step-by-step explanation:

The parameters given are;

Plant A:

Weeks, Height

0, 3

1, 4.8

2, 6.6

3, 8.4

The rate of change of height, H per week, t, $\left (\dfrac{dH}{dt} \right )$ for plant A per week is therefore;

$\dfrac{dH}{dt} = \dfrac{H_n - H_{(n-1)} }{t_n - t_{(n-1)}} = \dfrac{8.4 - 3 }{3 - 0} = \dfrac{5.4}{3} = 1.8 \ inches/week$

Therefore we have;

H = 1.8 × t + 3

At week 6,

H = 3 + 6×1.8 = 13.8 inches

Plant B

Weeks, Height

2, 7.3

3, 8.7

4, 10.1

The rate of change of height, H per week, t, $\left (\dfrac{dH}{dt} \right )$ for plant B per week is given as follows;

$\dfrac{dH}{dt} = \dfrac{H_n - H_{(n-1)} }{t_n - t_{(n-1)}} = \dfrac{10.1 - 7.3 }{4 - 2} = \dfrac{2.8}{2} = 1.4 \ inches/week$

Therefore we have;

When t = 2, H = 7.3 hence, 7.3 = 2 × 1.4 + H₀

Where:

H₀ = H at t = 0

H₀ = 7.3 - 2 × 1.4 = 4.5

At week 6 we have;

H = 4.5 + 6×1.4 = 12.9 inches

Which indicates that Plant A will be 0.9 inches taller than Plant B at week 6.

The correct option is therefore;

No, The greater rate of change of Plant A will result in it being 0.9 inches taller in 6 weeks.

5 0

3 years ago

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2 consecutive even numbers add to equal 178. What is the smallest of the 2 numbers?

Diano4ka-milaya [45]

Answer:

The smallest number is 88

Step-by-step explanation:

Let

x ----> the first consecutive even number

x+2 --->the second consecutive even number

we know that

The linear equation that represent this problem is given by

$x+(x+2)=178$

solve for x

$2x=178-2\\2x=176\\x=88$

so

$x=88\\x+2=88+2=90$

therefore

The smallest number is 88

4 0

3 years ago

hich statement best describes the domain and range of p(x) = 6–x and q(x) = 6x? p(x) and q(x) have the same domain and the same

Sati [7]

Answer:

$p(x)$ and $q(x)$ have the same domain and the same range.

Step-by-step explanation:

$p(x) = 6-x$ and

$q(x) = 6x$

First of all, let us have a look at the definition of domain and range.

Domain of a function $y =f(x)$ is the set of input value i.e. the value of $x$ for which the function $f(x)$ is defined.

Range of a function $y =f(x)$ is the set of output value i.e. the value of $y$ or $f(x)$ for the values of $x$ in the domain.

Now, let us consider the given functions one by one:

$p(x) = 6-x$

Let us sketch the graph of given function.

Please find attached graph.

There are no values of $x$ for which p(x) is not defined so domain is All real numbers.

So, domain is $(-\infty, \infty)$ or $x\in R$

Its range is also All Real Numbers

So, Range is $(-\infty, \infty)$ or $x\in R$

$q(x) = 6x$

Let us sketch the graph of given function.

Please find attached graph.

There are no values of $x$ for which $q(x)$ is not defined so domain is All real numbers.

So, domain is $(-\infty, \infty)$ or $x\in R$

Its range is also All Real Numbers

So, Range is $(-\infty, \infty)$ or $x\in R$

Hence, the correct answer is:

$p(x)$ and $q(x)$ have the same domain and the same range.

4 0

3 years ago

What initial investment must be made to accumulate $60000 in 17 years if the money is invested in a mutual fund that pays 12% an

mars1129 [50]

$7881.18

Step-by-step explanation:

Let the initial Investment be $P_{0}$ . The Interest is compounded on a monthly basis at 12% annual interest rate. After 17 years, the Investment amounts to $60,000.

As the annual interest rate is 12%, the monthly interest rate is 1%.

Since this is a compound interest problem, the total amount can be modeled as follows: $P(t)=P_{0}(1+\frac{i}{100})^{t}$

Here $i$ is the interest rate, i.e $1$ , and t is the number of time periods, i.e $17\textrm{ years x }12\frac{\textrm{months}}{\textrm{year}}$ = $204\textrm{ months}$

$60,000=P_{0}\textrm{ x }(\frac{101}{100})^{204}$

$P_{0}=7881.18$

∴ Initial Investment = $7881.18

4 0

3 years ago