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tankabanditka [31]
3 years ago
9

Show how to rewrite 1/5 with the denominator 10.

Mathematics
1 answer:
Gnoma [55]3 years ago
8 0

Answer:

first blank: 2

second blank: 2

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Please help me with this problem ASAP, Thanks!!<br>​
Tema [17]

Answer:

x=28

y= -4

Step-by-step explanation:

9x+6y= -12

-x-4y= -12

multiply -x-4y=-12 by 9 so the x's are equal

9x+6y= -12

-9x-36y= -108

add the equations so the x's cancel out

-30y=120

y= -4

plug y into an equation to find x

-x-4(-4)=-12

-x+16= -12

-x= -28

x=28

6 0
3 years ago
How many solutions does the equation below have?
miskamm [114]

Answer:

1. none

Step-by-step explanation:

When you simplify the expression, you see that the statement is is false for any value of x.

7 0
3 years ago
Read 2 more answers
Is -6, -38 a solution to y= 7x+4
Sunny_sXe [5.5K]

Answer:

Yes, (-6,-38) is a coordinate pair off the equation

Step-by-step explanation:

Plug y=7x+4 into your graphic caculator

press 2ND, GRAPH and scroll up to see the point.

4 0
3 years ago
Suppose that $5500 is placed in an account that pays 2% interest compounded each year.
pogonyaev

Answer: After 1 year:    $5,610

After 2 years: $5,722.20

Step-by-step explanation: Use the formula for periodic compounding interest, which is

A = P(1 + r/n)^(nt), where A is the final amount, P is the initial deposit, r is the interest rate as a decimal, n is the number of times the interest is compounded per year, and t is how many years.

Here, P = 5,500, r = 0.02  (that's 2% as a decimal), n = 1,

t = 1 for the first answer, t = 2 for the second answer (1 year, then for 2 years)

Plug the known values in to solve...

For 1 year...

A = 5,500(1 + 0.02/1)^(1*1)

  A = 5,500(1.02)^1

     A = 5,610

For 2 years...

A = 5,500(1 + 0.02/1)^(1*2)

  A = 5,500(1.02)²

     A = 5,722.20

3 0
2 years ago
Afirm works 7 days a week. Every employee must work exactly 3 full days and 4 half-days each week. A half-day can be either morn
Serhud [2]

Answer:

  a)  560 (as you know)

Step-by-step explanation:

There are 7C3 = 35 ways to choose 3 full-time days from 7.

There are 2^4 = 16 ways to choose a half-day each day from the remaining 4 days.

The total number of possible schedules is ...

  35×16 = 560

_____

nCk = n!/(k!(n-k)!)

7C3 = 7·6·5/(3·2·1) = 7·5 = 35

4 0
3 years ago
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