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Harrizon [31]
2 years ago
8

Solve three and two fifths plus three and four sixths

Mathematics
2 answers:
Rasek [7]2 years ago
4 0

Answer:

7 1/15

Step-by-step explanation:

3 2/5 + 3 4/6 = 7 1/15

Juli2301 [7.4K]2 years ago
4 0

Answer:

7 1/15

Step-by-step explanation:

  • 3\frac{2}{5} + 3\frac{2}{3}
  • \frac{17}{5} + \frac{11}{3}
  • \frac{51}{15} + \frac{55}{15}
  • \frac{106}{15}
  • 7\frac{1}{15}

Therefore, the answer is 7 1/15.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

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You are given two pieces of information: (1) the price of
mylen [45]

Answer:

c. divide the cars gas milelage

5 0
3 years ago
Applying properties of Exponents in exercise,use the properties of exponents to simplify the expression.See example 1.
evablogger [386]

Answer: a) 15625, b) 0.2, c) 625, d) 0.008.

Step-by-step explanation:

Since we have given that

a) (5^2)(5^3)

As we know that

a^m\times a^n=a^{m+n}

So, it becomes,

5^2\times 5^3=5^{2+3}=5^6=15625

(b) (5^2)(5^{-3})

So, it becomes,

5^2\times 5^{-3}=5^{2-3}=5^{-1}=\dfrac{1}{5}=0.2

(c) (5^2)^2

As we know that

(a^m)^n=a^{mn}

So, it becomes,

5^{2\times 2}=5^4=625

(d) 5^{-3}

a^{-m}=\dfrac{1}{a^m}\\\\5^{-3}=\dfrac{1}{5^3}=\dfrac{1}{125}=0.008

Hence, a) 15625, b) 0.2, c) 625, d) 0.008.

4 0
3 years ago
How do I do part c?
neonofarm [45]
Think you need to add everything up
6 0
3 years ago
Pls help!!!!!!!!!!!!!!
SVETLANKA909090 [29]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
A baseball player comes up to bat 3 times during a league game. He either gets a hit or gets an out. How many different combinat
Stella [2.4K]

Answer:

8 different combinations are possible.

Step-by-step explanation:

Here, we have 2 different combinations for each time.

And the player comes out to bat 3 times.

So, total number of combinations are:

2^{3} i.e. a total of 8 number of times.

Let a hit is termed as 'H' and an out is termed as 'O'.

Total combinations are:

{HHH, HHO, HOH, HOO, OHH, OHO, OOH, OOO}

Kindly have a look at the tree diagram attached in the answer area.

In starting, there are 2 combinations possible, i.e. 'O' and 'H'.

After 'O' , 2 possible i.e. 'O' and 'H'.

After 'H' , 2 possible i.e. 'O' and 'H'.

and so on....

8 0
3 years ago
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