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Brilliant_brown [7]
2 years ago
12

Jay has an album that holds 600 coins. Each page of the album holds 6 coins. If ​39% of the album is​ empty, how many pages are

filled with coins ​?
____ pages are filled with coins
Pls help me with this
Mathematics
1 answer:
Nutka1998 [239]2 years ago
6 0

9514 1404 393

Answer:

  61

Step-by-step explanation:

If there are 6 coins on each page, then the number of pages for 600 coins is ...

  (600 coins)/(6 coins/page) = 100 pages

__

If 39% are empty, then 100% -39% = 61% are filled.

The number of filled pages is ...

  61% × 100 = 61/100 × 100 = 61 . . . . pages filled

_____

<em>Comment about percent</em>

You can think of the percent symbol (%) and "per 100" (/100) as being interchangeable. 61% = 61/100

("Cent" is used to signify one of 100 parts. A cent is 1/100 of a dollar, for example. In land area, a cent is 1/100 of an acre. Then, "percent" can be thought of as "per cent" or "per 100 parts".)

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My name is Ann [436]
It's shifted two units to he left because: y=(x-h) means that the graph is shifted h units to the right and when the sign is positive you know that h is negative (subtracting a negative number is the same as adding the absolute value of that number )
4 0
3 years ago
Divide (-2x-3+8x^2) by (1 + 4x)
Strike441 [17]

Answer:

8x^2-2x-3/1+4x

Step-by-step explanation:

It doesn't say any of your answers but i just used a calculator and that's what i got

Sorry if its wrong

8 0
3 years ago
Direction: True or False
Goshia [24]

Answer:

<u>False </u>Any three points are always coplanar

<u>False </u>Two points are always collinear

<u>False </u>Two planes intersect at a point

Read explanations, as the answer to these questions, is subjective.

Step-by-step explanation:

The definition of coplanar is; figures that exist on the same plane. Any three points might not always be colinear, as some points might exist on one plane, but other points could exist on other planes. To visualize this phenomenon, refer to the attached image. Two points could be on the red plane, but the third could be on the green. Therefore, while there are three points, not all of them exist on the same plane. However, another plane can be constructed to connect these points, bear in mind certain problems might specifically indicate that three points are not coplanar, therefore, the answer is false.

The definition of collinear is existing on the same line. This statement is a little subject, it can be both true and false depending on the way one looks at it. A line can be drawn to connect any two points, one only needs two points to determine a line. So technically two points are always collinear. However, it also depends on the circumstance. Certain geometry problems might specifically indicate that two points are not collinear. Thus, it really depends on the context. Therefore, the answer is technically false.

Two planes usually intersect on a line. To understand this, please refer to the image attached. Tehcnailly, if a corner of one plane intersects another, then one can state that the plane intersects on a point, but typically, planes intersect on lines. Therefore, technically the answer to this problem is false, as two planes usually intersect on a line.

Images credits: Geogebra

8 0
3 years ago
Read 2 more answers
Express this to single logarithm
zzz [600]

Answer:  \log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

===========================================================

Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

  1. log(A) + log(B) = log(A*B)
  2. log(A) - log(B) = log(A/B)
  3. B*log(A) = log(A^B)

From there, we can then say the following:

\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}

8 0
3 years ago
Why is this so harddddd
qaws [65]

Answer:

there is no attachmengt

Step-by-step explanation:

8 0
2 years ago
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