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antiseptic1488 [7]
3 years ago
15

9/10-?=1/5 please help

Mathematics
2 answers:
zlopas [31]3 years ago
5 0
The answer is 7/10.              
anygoal [31]3 years ago
3 0
The answer is 7/10:
1/5 and 9/10 HAVE to have the same denominator for us to do this problem.
So: 5 and 10 have a multiplier in common: 2
So: 1/5 x 2/2 = 2/10
Then you subtract: 9/10 - 2/10 which equals to:
7/10!!!
Your welcome!!!
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Aaron needs a total of $410 to buy a new bicycle. He has $35 saved. He earns $15 each week delivering newspapers. How many weeks
Ostrovityanka [42]
35+15x=410
410-35=375
375/15=25
He will need to work 25 weeks.
3 0
3 years ago
HELP PLEASE ILL GIVE BRAINLIST 8th grade ALGEBRA I INSERTED A PHOTO
Alex73 [517]

Answer:

No

Step-by-step explanation:

Solve it yourself, or ask your mom. They might help. So yeah um bye

6 0
2 years ago
8) 7(p-6)= 8p - 34<br> any chance i can get the steps on how to solve this?
yan [13]

Answer:

Step-by-step explanation:

7(p-6)=8p-34

Distribute 7 into the parentheses on the Left Hand Side of the equation

7p-42=8p-34

get p together, so minus 7p on both sides

7p-7p-42=8p-7p-34

7p and -7p cancel out on L.H.S

-42=p-34

isolate p by adding 34 to both sides

-42+34=p-34+34

-34 and postitive 34 cancel out on the R.H.S

p=-8

Hope this helps! :)

8 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
What is 3.06 rounded to the nearest whole number
Gnom [1K]
3.06 rounded to the nearest whole number is 3.
.06 is much closer to 0 than it is to 1, so round down.
3.06 rounded down is 3, so that is your answer.
4 0
3 years ago
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