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2 years ago

Starla is measuring the floor of a rectangular room to determine how many square feet of a tile to buy to cover the floor

Mathematics

1 answer:

lbvjy [14]2 years ago

5 0

there isnt much here but answer is width=X and length=Y and X times Y

for example, if you had a room that was 3 feet long and 10 feet wide, it would be 30 square feet.

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If Patricia drives at two-third of her usual speed, she covers a certain distance in one hour more than the time she takes while

zalisa [80]

Let r = usual driving rate
let t = usual driving time
We need to figure out t

The distance she covers in her usual time at her usual rate is r*t

The distance she covers in her new time at her new rate is:
(1+t)*((2/3)r)

Set this equal to each other and solve for t.

rt = (2/3)r + (2/3)rt
(1/3)rt = (2/3)r
(1/3)t = (2/3)
t = 2

So her usual time is 2 hours. (There's probably a faster way to do this)

3 0

2 years ago

What is the recursive rule for an=4n−1? a1=4;an=an−1−1 a1=3;an=an−1+4 a1=−1;an=an−1+4 a1=3;an=an−1−1

navik [9.2K]

Plug in n = 1 into the nth term formula
a(n) = 4n-1
a(1) = 4*1-1
a(1) = 3
So the first term is 3

The second term will be 7 because we add on 4 each time, as indicated by the slope of 4. This is also known as the common difference.

So the nth term is found by adding 4 to the (n-1)st term, in other words,
a(n) = a(n-1)+4

----------------------------------------------------------------------------

In summary, the answer is
a1=3; an=an-1+4
which is choice B

4 0

2 years ago

Read 2 more answers

Find the range of y=3/2cos4x-1

MrRissso [65]

Answer:

Range = [- 2.5, 0.5] = [ - 5/2, 1/2]

Step-by-step explanation:

Smallest value of cos α = - 1,

largest value of cos α = 1.

When cos 4x = - 1, y=3/2cos4x-1 = 3/2*(-1) - 1 = - 5/2 = - 2 1/2 = - 2.5

When cos 4x = 1, y=3/2cos4x-1 = 3/2*1 - 1 = 1/2 = 0.5

Range = [- 2.5, 0.5] = [ - 5/2, 1/2]

5 0

2 years ago

Below is the five Number summary for a 136 hikers who recently completed the John muir trail JMT the variable is the amount of t

Helga [31]

Answer:

$IQR = Q_3 -Q_1 = 28-18 = 10$

Now we can find the limits in order to determine outliers like this:

$Left = Q_1 -1.5 IQR = 18 -1.5*10=3$

$Right = Q_1 +1.5 IQR = 18 +1.5*10=33$

So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier

b. 3

Step-by-step explanation:

For this case we have the following summary:

$Minimum = 9$ represent the minimum value

$Q_1 = 18$ represent the first quartile

$Median =Q_2= 21$ represent the median

$Q_3 = 28$ represent the third quartil

$Maximum=56$ represent the maximum

If we use the 1.5 IQR we need to find first the interquartile range defined as:

$IQR = Q_3 -Q_1 = 28-18 = 10$

Now we can find the limits in order to determine outliers like this:

$Left = Q_1 -1.5 IQR = 18 -1.5*10=3$

$Right = Q_1 +1.5 IQR = 18 +1.5*10=33$

So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier

b. 3

7 0

2 years ago

The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in

Marina86 [1]

Answer:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

$\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3$