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NeTakaya
2 years ago
10

Wendy is arranging books on the bookshelves in the school library. The total number of books she arranges is given by the equati

on b = 12r, where b is the total number of books and r is the number of rows. Each row contains the same number of books.
If Wendy arranged 16 rows, she arranged a total of
books.
Mathematics
2 answers:
horrorfan [7]2 years ago
7 0
You would do 16x12 which is 192

Basile [38]2 years ago
6 0
Given:
b = 12r 
where:
b =  represents the total number of books
12 = number of books per row
r = number of rows

If Wendy arranged 16 rows, she arranged a total of ___ books.

b = 12r
b = 12(16)
b = 192 total number of books arranged.
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X - 9 + 2wx = y Add 9 to both sides x + 2wx = y + 9 Factor out the x x (1 + 2w) = y + 9 Divide both sides by (1 + 2w) x = (y + 9) / (1 + 2w)
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Lizzy had test scores of: 72, 94, 108, 60What is the RANGE of her test scores?
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48

Step-by-step explanation:

To find the range, you have to subtract the biggest/largest number by the smallest number:

108 - 60 = 48

The range is 48.

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A bicycle manufacturing company makes a particular type of bike. Each child bike requires 4 hours to build and 4 hours to test.
Leokris [45]

Answer:a

Step-by-step explanation:

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2 years ago
Solve x2 = 12x – 15 by completing the square. Which is the solution set of the equation? (negative 6 minus StartRoot 51 EndRoot
Amanda [17]

Answer:

LAST OPTION: (6-\sqrt{21},6+\sqrt{21})

Step-by-step explanation:

1. Subtract 12x from both sides of the equation:

x^2-12x= 12x- 15-12x\\\\x^2-12x=-15

2. Since b=12:

(\frac{b}{2})^2=(\frac{12}{2})^2=(6)^2

3. Now can complete the square. Add (6)^2 to both sides of the equation:

x^2-12x+6^2=-15+6^2

4. Simplifying:

(x-6)^2=21

5. Solve for "x":

\sqrt{(x-6)^2}=\±\sqrt{21}\\\\x-6=\±\sqrt{21}\\\\x=6\±\sqrt{21}\\\\x_1=6+\sqrt{21}\\\\x_2=6-\sqrt{21}

6. The solution set is:

(6-\sqrt{21},6+\sqrt{21})

3 0
2 years ago
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The length of the day in Boulder (Latitude 40 N) can be modeled approximately by l(t) = −3 cos ( 2π 365 (t + 10)) + 12 where l i
Harman [31]

Answer:

(a) I(355)=9

(b) I'(265)=−0.05163

(c) The l′(t) is largest at 81.25.

Step-by-step explanation:

The given function is

I(t)=-3\cos (\frac{2\pi}{365}(t+10))+12

where I(t) the length of the day in Boulder and l is given in hours and t is the day of the year.

(a)

Substitute t=355 in given function.

I(355)=-3\cos (\frac{2\pi}{365}(355+10))+12

I(355)=-3\cos (2\pi)+12

I(355)=-3(1)+12

I(355)=9

Therefore, the value of I(355) is 9. It means the length of the day in Boulder is 9 hours at 355 day of the year.

(b)

Differentiate the given function with respect to t.

I'(t)=-3(-\sin (\frac{2\pi}{365}(t+10)))(\frac{2\pi}{365})

I'(t)=\frac{6\pi}{365}\sin (\frac{2\pi}{365}(t+10))

Substitute t=265 in the above function.

I'(t)=\frac{6\pi}{365}\sin (\frac{2\pi}{365}(265+10))

I'(265)\approx −0.05163

Therefore, the value of I'(265) is −0.05163. It means the length of the day in Boulder decreased by 0.05163 hours at 265 day of the year.

(c)

Differentiate the I'(t) function with respect to t.

I''(t)=\frac{6\pi}{365}\cos (\frac{2\pi}{365}(t+10))(\frac{2\pi}{365})

I''(t)=\frac{12\pi^2}{(365)^2}\cos (\frac{2\pi}{365}(t+10))

Equate I''(t)=0 to find critical points.

\frac{12\pi^2}{(365)^2}\cos (\frac{2\pi}{365}(t+10))=0

\cos (\frac{2\pi}{365}(t+10))=0

\cos (\frac{2\pi}{365}(t+10))=\cos (\frac{\pi}{2})

On comparing both sides we get

\frac{2\pi}{365}(t+10)=\frac{\pi}{2}

t+10=\frac{\pi}{2}\times \frac{365}{2\pi}

t+10=\frac{365}{4}

t=\frac{365}{4}-10

t=81.25

Check the value of the function I'''(t) at t=81.25.

Since I'''(81.25)=-0.0000153, therefore the value of l′(t) is largest at 81.25.

4 0
2 years ago
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