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DanielleElmas [232]
3 years ago
9

I need help -6x > 30

Mathematics
1 answer:
Debora [2.8K]3 years ago
7 0

Answer:

×< -5

Step-by-step explanation:

-6x>30 reverse the sign

x< -5

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Which number can each term of the equation be multiplied by to eliminate the decimals before solving?
Sati [7]
Ok...

5.6 = 56/10 = 560/100

1.1 = 11/10 = 110/100

0.12 = 12/100

--------------

\frac { 560 }{ 100 } j-\frac { 12 }{ 100 } =4+\frac { 110 }{ 100 } j\\ \\ \\ 100\times \left( \frac { 560 }{ 100 } j-\frac { 12 }{ 100 }  \right) =\left( 4+\frac { 110 }{ 100 } j \right) \times 100\\ \\ 560j-12=400+110j\\ \\ 560j-110j=400+12\\ \\ 450j=412\\ \\ j=\frac { 412 }{ 450 }

So, the answer is: 100

You could multiply both sides of the equation by 100 to get the value of (j) quickly.
6 0
3 years ago
Read 2 more answers
What is the square root of -1?
Minchanka [31]
The square root of -1 “i”
5 0
3 years ago
Write the equation for:
xenn [34]

Answer:

\frac{1}{2} x +x = 2 (x-4)

Step-by-step explanation:

Suppose number is x

Following equation can be written from question statement

\frac{1}{2} x +x = 2 (x-4)

Simplifying it further

\frac{3}{2} x = 2x-8

\frac{3}{2} x -2x=-8

\frac{3x-4x}{2}= -8

-x= -16\\x=16

7 0
3 years ago
Let HH represent the number of hummingbirds and SS represent the number of sunbirds that must pollinate the colony so it can sur
Nikitich [7]

Answer: 7

Step-by-step explanation:

Here the given inequality that shows the number of humming birds(H) and the number of sunbirds(S)  that must pollinate the colony so it can survive until next year,

6 H + 4 S > 74

If This year, 8 hummingbirds pollinated the colony.

By putting H=8 in the above inequality,

We get,

6× 8 + 4 S > 74

⇒ 48 + 4 S > 74

⇒ 4 S > 26

⇒ S > 6.5

Hence the least number of sunbirds is 7.

7 0
3 years ago
Read 2 more answers
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
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