Assuming you mean y = 200 - 16t^2, we have all the required information needed to solve this problem. The y-value is the height of the building and the t-value represents the number of seconds after the shoe fell off.

Since we are trying to solve <em>for t</em>, we will be using our y-value. The problem states that we landed on a building with a height of 31 feet. We can plug this into the y-value, since that is what y is defined as (the height of the building).

Now we have:
31 = 200 - 16t^2

We can solve this to find t:
-169 = -16t^2
169 = 16t^2
10.5625 = t^2
3.25, -3.25 = t

We have found two answers for t. However, the negative value is not a solution because we can not have a negative number of seconds. Thus, 3.25 is the only value of t which works for this problem.

Since t is the value we are trying to find, we have our answer. The shoe hits the building after 3.25 seconds.

7 0

3 years ago

Read 2 more answers

Determine the y-intercept and slope and then write the equation for this table.*

Novosadov [1.4K]

Answer:

The y-intercept is 15 and the slope is 5.5

y=5.5x+15

Step-by-step explanation:

4 0

3 years ago

Find the sum of the arithmetic sequence 1, 3, 5, 7, … , 99.

bezimeni [28]

The sum of the arithmetic sequence 1, 3, 5, 7, … , 99 is 2525

<h3>What is Sequence?</h3>

Sequence is an enumerated collection of objects in which repetitions are allowed and order matters.

Given sequence is 1, 3, 5, 7, … , 99.

It is a arithmetic sequence, the difference between two consecutive terms will be same

d=3-1=2

first term a=1

l=last term =99

We need to find 99 is which term in the sequence

Aₙ = a + ( n - 1 ) d

99=1+(n-1)2

99=1+2n-2

99=2n-1

100=2n

Divide both sides by 2

50=n

Now we need to find the sum of 50 terms

Sₙ=n/2(l+d)

=50/2(99+2)

=25(101)

=2525

Hence, the sum of the arithmetic sequence 1, 3, 5, 7, … , 99 is 2525

To learn more on Sequence click:

brainly.com/question/21961097

#SPJ1

6 0

1 year ago

The weight of Jane and Jessica was in the ratio of 8 : 9. Jane gained 2 kg while Jessica lost 4 kg. They then had the same weigh

postnew [5]

Answer:

48 kg

Step-by-step explanation:

The given relations can be used to write a system of equations for the two weights. Those can be solved to find Jane's weight.

<h3>setup</h3>

Let x and y represent Jane's and Jessica's original weight, respectively. The ratio of weights was ...

x/y = 8/9

After the changes in weight, they were equal:

x+2 = y-4

<h3>solution</h3>

Adding 4 to the second equation, we have an expression for y that can be substituted into the first equation.

y = x +6 . . . . . . . . . . solve the second equation for y

x/(x+6) = 8/9 . . . . . . substitute for y in the first equation

9x = 8(x +6) . . . . . . . multiply by 9(x+6)

x = 48 . . . . . . . . . . . simplify and subtract 8x

Jane weighed 48 kg at first.

_____

<em>Alternate solution</em>

The original difference in "ratio units" was 9-8 = 1 ratio unit. We find that this corresponds to 6 kg after the weight changes make the weights equal. Then 8 ratio units will be 8(6 kg) = 48 kg—Jane's original weight.

(This mental solution is virtually the same as the solution using equations shown above.)

3 0

1 year ago