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Setler79 [48]
3 years ago
9

The sum of two numbers is 70. the first number is 2/5 of the second number. What are the numbers?

Mathematics
2 answers:
11111nata11111 [884]3 years ago
3 0
A + b = 70
a = 2/5b

2/5b + b = 70
2/5b + 5/5b = 70
7/5b = 70
b = 70 * 5/7
b = 350/7
b = 50

a = 2/5b
a = 2/5(50)
a = 100/5
a = 20

the numbers are 50 and 20....the smallest being 20 and the largest being 50


dolphi86 [110]3 years ago
3 0
X + 2/5x = 70
1x + 2/5x = 70
1 2/5 x = 70
7/5 x = 70  multiply both sides by 5
7x = 350  divide both sides by 7
x = 50    so 2/5 x = 20    the two numbers are 50 and 20
the larger of the two is 50
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6 0
3 years ago
Read 2 more answers
What is the sum (2/5x + 5/8) + (1/5x - 1/4)? a. 3/5 x + 1/8 b. 3/5 x + 3/8 c. 3/5x + 5/8 d. 3/5x + 7/8​
ozzi

Answer:

3/5x + 3/8

Step-by-step explanation:

2/5x + 5/8 + 1/5x - 1/4

1. Combine like terms so now u have

3/5x + <u><em>3/8 </em></u>

In case you want to know how I got 3/8

You want a common denominator so

from  5/8 + 1/4, make both denominator 8

so 1 *2/ 4* 2

so now you converted the 1/4 to 2/8

So then 5/8 - 2/8 gives u 3/8 and that's how i got 3/8

Hope this helped!

6 0
3 years ago
Read 2 more answers
A collection of dimes and quarters is worth $6.50. There are 35 coins in all. How many dimes are there?
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Lets do this like this:
X = number of quarters and y=number of dimes.
Now, we know that the total number is 35 and that the change adds up to 6.50.
One equation we can form is <span>x+y=35</span>Remember the total number of coins is 35.
The other equation is <span>25x+.1y=6.5</span>Because the number of dimes times their value plus the number of quarters times their value gives us a total of 6.5. Now we can solve for x or y easily in the first equation.
Solving for y like this:<span>y=35−x</span>We can substitute this into the other equation to obtain<span>.25x+.1(35−x)=6.5</span>This simplifies to<span>.15x+3.5=6.5</span><span>.15x=3</span><span>x=<span>20
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8 0
3 years ago
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Natasha_Volkova [10]

Answer:

x = 15

Step-by-step explanation:

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3 0
3 years ago
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