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Sedaia [141]
3 years ago
7

Help me please and thank you

Mathematics
1 answer:
Yuri [45]3 years ago
4 0

Answer:

A

Step-by-step explanation:

Solution is in this picture.

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What is the answer for this equation 600 is 4/9 of what number 
horrorfan [7]

Answer:

933.3333333

Step-by-step explanation:

6 0
2 years ago
9/10 divided by 3/5<br> please helppp
Westkost [7]

Answer:

=1.5

Step-by-step explanation:

3 0
3 years ago
If C = 2m2 + m and D = 2 – 6m + 2m², find an expression that equals<br> 2C - 2D in standard form.
lidiya [134]
C = 2m^2 + m
D = 2 - 6m + 2m^2

2C = 2(2m^2 + m) = 4m^2+2m
2D = 2(2-6m+2m^2) = 4-12m+4m^2

2C - 2D =
4m^2+2m-(4-12m+4m^2) =
4m^2+2m-4+12m-4m^2 =
0m^2 + 14m -4 =
14m - 4
3 0
3 years ago
How many of the numbers from 10 through 92 have the sum of their digits equal to a perfect​ square?
horrorfan [7]
All the numbers in this range can be written as 10d_1+d_0 with d_1\in\{1,2,\ldots,9\} and d_2\in\{0,1,\ldots,9\}. Construct a table like so (see attached; apparently the environment for constructing tables isn't supported on this site...)

so that each entry in the table corresponds to the sum of the tens digit (row) and the ones digit (column). Now, you want to find the numbers whose digits add to perfect squares, which occurs when the sum of the digits is either of 1, 4, 9, or 16. You'll notice that this happens along some diagonals.

For each number that occupies an entire diagonal in the table, it's easy to see that that number n shows up n times in the table, so there is one instance of 1, four of 4, and nine of 9. Meanwhile, 16 shows up only twice due to the constraints of the table.

So there are 16 instances of two digit numbers between 10 and 92 whose digits add to perfect squares.

7 0
3 years ago
Use limits to find the area of the region bounded by the graph f(x)=4-2x^3 , the x-axis , and the vertical lines x=0 and x=1
gtnhenbr [62]

Answer:

\frac{7}{2} square units.

Step-by-step explanation:

We have to use limits to find the area of the region bounded by the graph f(x) = 4 - 2x^{3} , the x-axis, and the vertical lines x=0 and x=1.

So, the area will be

A = \int\limits^1_0 {(4 - 2x^{3})} \, dx

= [4x - \frac{x^{4}}{2} ]^{1} _{0}

= 4 - \frac{1}{2}

= \frac{7}{2} square units. (Answer)

3 0
3 years ago
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