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love history [14]
4 years ago
12

9 times as much as 7 tenths

Mathematics
2 answers:
Luba_88 [7]4 years ago
8 0

9 times as much of 7 things is 9 x 7 = 63 things, so 9 times as much as 7 <em>tenths </em> will be 9 x 7 = 63 <em>tenths.</em>

tamaranim1 [39]4 years ago
3 0

Answer:

The value of 9 times as much as 7 tenths is 6.3.

Step-by-step explanation:

We need to find the number that is 9 times as much as 7 tenths.

The value of 7 tenths is \frac{7}{10}.

9 times as much as 7 tenths would be

9\times \frac{7}{10}

Now we have to evaluate the value of the above expression.

\frac{9\times 7}{10}

\frac{63}{10}

On further simplification we get

6.3

Therefore, the value of 9 times as much as 7 tenths is 6.3.

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Aneli [31]

Answer:

a. 450 b.  10078.9097

Step-by-step explanation:

a. go with the formula W=FD

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3 years ago
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Flauer [41]

Answer:

x = 10

Step-by-step explanation:

50 = (6x - 10)      rewrite it

50 =  6x - 10

+ 10        + 10       add 10 to both sides

60 =  6x               you're left with 60 and 6x

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7 0
3 years ago
Johnny and sally bought 4 pounds of mulch and two pounds if seedfor the garden.how many ounces of these products did they purcha
iren2701 [21]
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8 0
3 years ago
HELP!!!
ella [17]

Answer:

120

Step-by-step explanation:

We are given that the function for the number of students enrolled in a new course is f(x) = 4^{x}-1.

It is asked to find the average increase in the number of students enrolled per hour between 2 to 4 hours.

We know that the average rate of change is given by,

A = \frac{f(x)-f(a)}{x-a},

where f(x)-f(a) is the change in the function as the input value (x-a) changes.

Now, the number of students enrolled at 4 = f(4) = f(x) = 4^{4}-1 = 255 and the number of students enrolled at 2 = f(2) = f(x) = 4^{2}-1 = 15

So, the average increase A=\frac{f(4)-f(2)}{4-2} = A=\frac{255-15}{4-2} = A=\frac{240}{2} = 120.

Hence, the average increase in the number of students enrolled is 120.

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4 years ago
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What type of problem is it?
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