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Nesterboy [21]
3 years ago
10

Freddy has a shelf that is 72 over 5 inches wide. How many catalogs can Freddy arrange on the shelf if each catalog is 4 over 5

of an inch thick? (5 points)
a
11, because the number of catalogs is 72 over 5 divided by 5 over 4

b
12, because the number of catalogs is

c
16, because the number of catalogs is 72 over 5 multiplied by 4 over 5

d
18, because the number of catalogs is 72 over 5 divided by 4 over 5
Mathematics
2 answers:
Amanda [17]3 years ago
7 0

d:
18, because the number of catalogs is 72 over 5 divided by 4 over 5
scZoUnD [109]3 years ago
5 0
D
Step-by-step explanation:
Let Freddy could arrange n catalogs on the shelf.
The dimension of the shelf is:
72 over 5 inches wide.
The dimension of the catalog is:
4 over 5 of an inch thick.
n × Dimension of one catalog=Dimension of shelf.
i.e.
n×(4 over 5 inches)=72 over 5 inches
i.e.
n=72 over 5 inches/ 4 over 5 inches
i.e.
n=72/4
i.e.
n=18
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Licemer1 [7]

Answer:

<em>No values of x can make f(x)=6</em>

Step-by-step explanation:

Equation with Absolute Value

The absolute value of a number is always positive. That condition must be met when solving equations. Any condition that goes against the rule, must be discarded and not part of the solution.

The function provided in the question is:

f(x) = -0.5|2x + 2| + 1

We need to find the value(s) of x that make:

f(x)=6

It needs to solve the equation:

-0.5|2x + 2| + 1=6

Subtracting 1:

-0.5|2x + 2| =6-1=5

Dividing by -0.5:

|2x + 2| =5/(-0.5)=-10

We reach to this equation to solve:

|2x + 2| =-10

As stated above, the absolute value is always positive, and the equation forces the absolute value to be negative. There is no possible value of x that makes the absolute value negative, thus:

No values of x can make f(x)=6

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dem82 [27]

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Here, you're mixing scores 85 and 90 with weights X and Y, respectively. You are asked for the ratio of X to Y.

There's a quick way to work mixture problems of all kinds. Write the two components of the mixture on the left. Here, they are 90 and 85. (I usually put the larger one on top.) Put the mixed value in the middle, and form differences along the lines of an X, as shown. The numbers on the right give the relative contributions of the constituents at the same level in the diagram. Here, the ratio of X to Y is shown as 2 to 3.

For some mixture problems, you need to know the proportion of the constituent to the whole. In that case, add the ratio values to get the "whole". For example, here the X class students make up 2/(2+3) = 2/5 of the whole number of students.

For your problem, X/Y = 2/3, corresponding to selection D.

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3 years ago
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