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son4ous [18]
4 years ago
12

The sum of 1/3 and twice a number is equal to 2/3 subtracted from three the number.find the number

Mathematics
1 answer:
Zigmanuir [339]4 years ago
4 0
The number = y

2y + 1/3 = 2/3 - 3y
(add 3y to both sides)
5y + 1/3 = 2/3
(subtract 1/3 from both sides to isolate y)
5y = 1/3
(divide by 5)
y = 1/15
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When an exponential expression is raised to another exponent the exponent is______​
goldenfox [79]

Answer:

Step-by-step explanation:

When raising a power to a power in an exponential expression, you find the new power by multiplying the two powers together. For example, in the following expression, x to the power of 3 is being raised to the power of 6, and so you would multiply 3 and 6 to find the new power.

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3 years ago
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Can you please help with math
melomori [17]
Interpret the exponent in the formula. S to the third power means to use s as a factor 3 times.

 V = S to the third power = S x S x S

Find the volume of the first cube.
Substitute 4 for s.

V = 4 x 4 x 4 = 64

Find the volume of the second cube.
Substitute 7 for s.

V = 7 x 7 x 7 = 343

Tamara's first cube has a volume of 64 cubic centimeters, and her second cube has a volume of 343 cubic centimeters 
7 0
3 years ago
What is the answer to<br> 3n=n-2
Aloiza [94]

Answer:

- 1

Step-by-step explanation:

Step 1:

3n = n - 2

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n = - 1

Hope This Helps :)

5 0
3 years ago
$119 croquet set 50% discount
natta225 [31]
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3 years ago
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|x-3|&lt;2<br> |4x+1|&gt;0<br> |x-1|&lt;5<br> Ayudaaaaaaa pleaseeeee
Dafna11 [192]

Recuerda que

• |<em>x</em>| = <em>x</em> si <em>x</em> ≥ 0

• |<em>x</em>| = -<em>x</em> si <em>x</em> < 0

Necesitas considerar dos casos:

• si <em>x</em> - 3 ≥ 0,

|<em>x</em> - 3| < 1   ⇒  <em>x</em> - 3 < 1   ⇒   <em>x</em> < 4

• si <em>x</em> - 3 < 0,

|<em>x</em> - 3| < 1   ⇒   -(<em>x</em> - 3) = 3 - <em>x</em> < 1   ⇒   -<em>x</em> < -2   ⇒   <em>x</em> > 2

Entonces la solución consta de todos los números reales <em>x</em> tales que <em>x</em> > 2 y <em>x</em> < 4, o simplemente 2 < <em>x</em> < 4.

El método para resolver las otras desigualdades es el mismo.

|4<em>x</em> + 1| > 0   ⇒   4<em>x</em> + 1 > 0   o   -(4<em>x</em> + 1) > 0

…   ⇒   4<em>x</em> + 1 > 0   o   -4<em>x</em> - 1 > 0

…   ⇒   4<em>x</em> > -1   o   -4<em>x</em> > 1

…   ⇒   <em>x</em> > -1/4   o   <em>x</em> < -1/4

⇒   <em>x</em> ≠ -1/4

|<em>x</em> - 1| < 5   ⇒   <em>x</em> - 1 < 5   o   -(<em>x</em> - 1) < 5

…   ⇒   <em>x</em> - 1 < 5   o   -<em>x</em> + 1 < 5

…   ⇒   <em>x</em> < 6   o   -<em>x</em> < 4

…   ⇒   <em>x</em> < 6   o   <em>x</em> > -4

⇒   -4 < <em>x</em> < 6

7 0
3 years ago
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