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Mandarinka [93]
3 years ago
12

What is a word problem for 5x + 10 ≥ 50

Mathematics
1 answer:
aliya0001 [1]3 years ago
7 0

solve for x by simplifying both sides of the equation,then isolating the variable x>8


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Jack has 9 flowers to plant. He plants 2 flowers before lunch. Then he plants 3 more after lunch. How many flowers does Jack hav
Vika [28.1K]

Answer:

4 flowers

Step-by-step explanation:

Number of flowers planted: 3+ 2 =5

Jack has 9 flowers that he needs to plant. We need to subtract 5 from 9 to find out how many flowers he still needs to plant.

9-5=4

Jack has 4 more flowers left to plant.

4 0
3 years ago
When a number is decreased by 4%, the result is 93. What is the original number
romanna [79]

Answer:

<u>The original number is 96.875</u>

Step-by-step explanation:

Let's find the original number using the Direct Rule of Three, this way:

       Percentage          Number

              100                       x

               96                       93

***********************************************

96x = 93 * 100

96x = 9,300

x = 9,300/96

x = 96.875

<u>The original number is 96.875</u>

7 0
3 years ago
-3 is greater than w , and -8 is less than w
saveliy_v [14]

Answer:

-8<w<-3

Step-by-step explanation:

7 0
4 years ago
beth has $80 to spend at mall. she bought 3 movies for $13 each and 1 book for $11. before she left she got a soda that was $3 .
OlgaM077 [116]
She have $47 left back.
8 0
3 years ago
Read 2 more answers
Use pascal's triangle to expand the following binomial expression <br>1.(2k-1/3)⁶​
Gwar [14]

9514 1404 393

Answer:

  64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

Step-by-step explanation:

The row of Pascal's triangle we need for a 6th power expansion is ...

  1, 6, 15, 20, 15, 6, 1

These are the coefficients of the products (a^(n-k))(b^k) in the expansion of (a+b)^n as k ranges from 0 to n.

Your expansion is ...

  1(2k)^6(-1/3)^0 +6(2k)^5(-1/3)^1 +15(2k)^4(-1/3)^2 +20(2k)^3(-1/3)^3 +...

     15(2k)^2(-1/3)^4 +6(2k)^1(-1/3)^5 +1(2k)^0(-1/3)^6

  = 64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

7 0
3 years ago
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