A swimming pool is 30 feet wide and 50 feet long. If it is surrounded by square tiles, each of which is 1 foot by 1 foot, how many tiles are there surrounding the pool?

Answers

Answer 1

Answer:

164 tiles

Step-by-step explanation:

Per the above details, we will require tiles along all the sides of the swimming pool ,hence the circumference of the pool is denoted as;

= 2(wide + long)

Wide = 30 feet

Long = 50 feet

Circumference = 2(30 + 50)

Circumference = 2(80)

Circumference = 160.

It means that minimum of 160 tiles of 1 ft length is required, however, we will also require tile on each of the corners(4), total equals

= 160 + 4

= 164 tiles

A total of 164 tiles are there surrounding the pool.

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Which of the following values "completes the square," or creates a perfect square trinomial, for x2 − 8x + ___?816–8–16
Which is equal to the product of HJ?

Which represents the solution(s) of the system of equations, y = –x2 + 6x + 16 and y = –4x + 37? Determine the solution set algebraically.a.(3, 25)
b.(–3, 49)
c.(3, 25) and (7, 9)
d.(–3, 49) and (–7, 65)

Answers

Answer: c.(3, 25) and (7, 9)

y = –x^2 + 6x + 16 and y = –4x + 37

Plug in -4x+37 for y in first equation . It becomes

$-x^2 + 6x + 16= -4x+37$

Combine like terms. add 4x and subtract 37 on both sides

$-x^2 + 10x - 21=0$

Divide the whole equation by -1 to remove negative sign from -x^2

$x^2 - 10x + 21=0$

Now factor the left hand side

(x-7)(x-3) = 0

x-7 =0 and x-3=0

x= 7 and x=3

Now we find out y using y = –4x + 37

when x= 7 , then y=-4(7) +37 = 9

when x= 3, then y=-4(3) + 37 = 25

We write solution set as (x,y)

(7,9) and (3,25) is our solution set

TO determine the solution set of the equations given above, we can use substitution method. We do as follows:

y = –x2 + 6x + 16 and y = –4x + 37

–x2 + 6x + 16 = –4x + 37
-x^2 +10x -21 = 0
x = 7 and 3
y = 9 and 25

Therefore, option C is the correct answer.

Please help! 25 points!

Answers

Step-by-step explanation:

NOT list 1 there is only one name posted

NOT list 2 some of the results have only one name

YES list 3 it has two names per result and some of the names repeat...consistent with drawing and replacement

NOT list 4 it has THREE names listed....there is only two drawings

Which of the following could not be true for a function?Domain is {2}, Range is {2}
Domain is {2, 3}, Range is {2}
Domain is {2}, Range is {2, 3}
Domain is {2, 3}, Range is {2, 3}

Answers

C. hope this helps :::)

Answer:

I agree with C

Step-by-step explanation:

11 5 over 8 into an improper fraction

Answers

11 5/8 as a improper fraction

1) Simplify Fraction
⇒ Multiply denominator with whole number
8 x 11 = 88

⇒ Answer of multiplication and add the numerator
88 + 5 = 93

93/8 is improper Fraction.
The fraction cant be simplified

11 5/8
11 x 8 = 88 We multiply the whole number by the denominator
88 + 5 = 93 We take the answer (88) and add that to 5, which equals 93.
The denominator stays the same, so you just place the 93 over 8
A= 93/8

Eliza Savage received a statement from her bank showing a checking account balance of $324.18 as of January 18. Her own checkbook shows a balance of $487.38 as of January 29. The bank returned all of the cancelled checks but three. The amounts of these three checks are $15.00, $77.49, and $124.28. How much did Eliza deposit in her account between January 18 and January 29?

Answers

Assuming that the three checks that were returned were deposited before Jan 29 and was returned also before Jan 29, the three deposited checks will be added to get 216.77. Next, we subtract 324.18 with 487.38 to get 163.2 which is the difference of Jan 18 and Jan29's balance. Then add 163.2 and 216.77 to get 379.97, the total amount deposited

Find the third directional cosine (there are two of them with opposite signs) if it is known that two directional cosines are equal to 1/2 and 1/3

Answers

Answer:

To find the third directional cosine, we need to use the property that the sum of the squares of the directional cosines is equal to 1.

Let's denote the three directional cosines as cosα, cosβ, and cosγ. Given that two directional cosines are equal to 1/2 and 1/3, we can set up the following equations:

cosα = 1/2

cosβ = 1/3

To find cosγ, we can rearrange the equation for the sum of the squares of the directional cosines:

cos²α + cos²β + cos²γ = 1

Substituting the given values, we have:

(1/2)² + (1/3)² + cos²γ = 1

Simplifying the equation, we get:

1/4 + 1/9 + cos²γ = 1

To solve for cos²γ, we can combine the fractions:

(9/36) + (4/36) + cos²γ = 1

(13/36) + cos²γ = 1

Now, we can solve for cos²γ by subtracting 13/36 from both sides:

cos²γ = 1 - 13/36

cos²γ = 23/36

Taking the square root of both sides, we find:

cosγ = √(23/36)

Since we are looking for the third directional cosine, there are two possible solutions, one positive and one negative. So, the third directional cosine can be either √(23/36) or -√(23/36).

In conclusion, the third directional cosine, cosγ, is either √(23/36) or -√(23/36).

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