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Hatshy [7]
2 years ago
13

Name 3 types of rational numbers you can plot on a number line.

Mathematics
2 answers:
balu736 [363]2 years ago
5 0
Anything above zero anything negative from zero and 0 itself
adelina 88 [10]2 years ago
4 0

Answer:

  • Center of the number line (origin)
  • Positive rational numbers are marked on the right side of zero
  • Negative rational numbers are marked on the left side of the zero
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A music service chargers a $2.99 monthly membership fee plus $0.05 for each song purchased. If Naomi charges for the month was $
8_murik_8 [283]

Answer:

158

Step-by-step explanation:

10.89-2.99 = 7.90

.5/7.90=158

3 0
3 years ago
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Miko and Raj get the same salary. Every month Miko saves
navik [9.2K]

Answer:

Step-by-step explanation:

The salaries are the same. Only the fraction of how much is saved is quite different.

Equation

x ( 9/11 - 7/9) = 1584

Explanation: The salary amount is x. The fractional amounts saved are also given. You have to take the difference because one of the people saves more than the other.

The amount that is more is 1584.

Solution

9/11 = 0.8182

7/9 = 0.7778

9/11 - 7/9 = 0.8182 - 0.7778 = 0.0404

Put that into the equation.

x(0.0404) = 1584                     Divide by 0.0404)

x(0.0404) = 1584/0.0404        

answer:x = 39204

8 0
2 years ago
Solve for w in P = 2w + 2l, if P = 38 and l = 12.
vova2212 [387]
W= 7

38=2w+2(12)
                /
38=2w + 24
-24        -24

14=2w
divid by 2    divid by 2

7=w
4 0
3 years ago
Read 2 more answers
(2x+3)²=125 find x<br>with full solution please​
Schach [20]

Answer:

<em>x = ± ( 5√5 - 3 ) / 2</em>

Step-by-step explanation:

We are given the equation ( 2x + 3 )² = 125;

( 2x + 3 )^{ 2 }  = 125 - Take Square Root on Either Side,\\| 2x + 3 | = \sqrt{ 125 } - Simplify Square Root of 125,\\\\| 2x + 3 | = \sqrt{ 5 * 25 } - Simplify Further,\\| 2x + 3 | = 5\sqrt{ 5 } - Take | |,\\\\2x + 3 = 5\sqrt{ 5 } , and , 2x + 5 = - 5\sqrt{ 5 } - Algebra, \\2x = 5\sqrt{ 5 } - 3 , and , 2x = - 5\sqrt{ 5 } - 3 - Divide Either Side by 2\\\\x = ( 5\sqrt{ 5 } - 3 ) / 2, and , x = ( - 5\sqrt{ 5 } - 3 ) / 2\\Conclusion ; x = ( + / - 5\sqrt{ 5 } - 3 ) / 2

<em>Solution ; x = ± ( 5√5 - 3 ) / 2</em>

7 0
3 years ago
How can I derive the first equation to get the second equation?
Jlenok [28]
Keeping in mind that "R" and "r" are constants whilst "s" is a variable and θ is a function in terms of "s", thus

\bf s^2=R^2+r^2-2Rrcos(\theta )\implies 2s^1=0+0-2Rr\left[ \stackrel{chain~rule}{-sin(\theta )\cfrac{d\theta }{ds}} \right]&#10;\\\\\\&#10;2s=2Rrsin(\theta )\cfrac{d\theta }{ds}\implies 2s=\cfrac{2Rrsin(\theta )d\theta }{ds}\implies \cfrac{2s\cdot ds}{2Rr}=sin(\theta )d\theta &#10;\\\\\\&#10;\cfrac{s\cdot ds}{Rr}=sin(\theta )d\theta
6 0
3 years ago
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