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pishuonlain [190]
3 years ago
6

Convert the recurring decimal 0.23333 into a fraction

Mathematics
2 answers:
bazaltina [42]3 years ago
8 0
x=0.23333...
\\
\\10x=10 \times 0.23333...
\\10x=2.33333...
\\
\\100x=100 \times 0.23333...
\\100x=23.3333...
\\
\\100x-10x=23.3333...-2.3333...
\\90x=21
\\x= \frac{21}{90}= \frac{7}{30}
Darina [25.2K]3 years ago
5 0
It would be: 0.23333 = 23333/100000
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yuradex [85]
Create equations:
10X+15y=2575.
X+Y=210.
Solve for X: X=210-Y, and sub it into the other equation: 10(210-Y)+15Y=2575. Solve for Y: 2,100-10y+15y=2575, 5y=475, y=95.
Sub Y into the original problem: X+95=210. Solve: X=115.
The first mechanic charges $115 per hour and the second charges $95 :)
6 0
3 years ago
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Two​ joggers, one averaging 8 mph and one averaging 4 ​mph, start from a designated initial point. The slower jogger arrives at
BaLLatris [955]
22.5 miles long


basically all you have to do is count out the factors until you reach two number of 45 apart

DIFFERENCE
7MPH 5MPH BETWEEN THEM

7 5 2
14 10 4
21 15 6
28 20 8
35 25 10
42 30 12
49 35 14
56 40 16
63 45 18
70 50 20
77 55 22
84 60 24
91 65 26
98 70 28
105 75 30
112 80 32
119 85 34
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133 95 38
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147 105 42
154 110 44
161 115 46



considering the differences are even it means it has to fall in the middle. if you count from the top it is 22.5 numbers down giving you the number of miles.
I'm sure there is an easier way but this is how i would solve it. :)
7 0
3 years ago
Write the sentence as an inequality <br><br> The sum of twice a number n and 8 is at most 25
Marta_Voda [28]

2n  + 8  \leqslant 25
5 0
4 years ago
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Sophie [7]

Answer:

can you explain what to do here?

Step-by-step explanation:

6 0
3 years ago
What are the points on y-axis which are at a distance of 4 units from 4x+3y=12
marissa [1.9K]

Given:

The equation of line is

4x+3y=12

To find:

The points on y-axis which are at a distance of 4 units from the given line.

Solution:

The point lie on the y-axis. So, their x-coordinate must be zero.

Let the points are in the form of (0,k).

The distance between a point (x_0,y_0) and a line ax+by+c=0 is

d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}

The given equation can be written as

4x+3y-12=0

The distance between (0,k) and 4x+3y-12=0 is 4 units.

4=\dfrac{|4(0)+3(k)-12|}{\sqrt{4^2+3^2}}

4=\dfrac{|0+3k-12|}{\sqrt{16+9}}

4=\dfrac{|3k-12|}{\sqrt{25}}

4=\dfrac{|3k-12|}{5}

Multiply both sides by 5.

20=|3k-12|

\pm 20=3k-12

12\pm 20=3k

\dfrac{12\pm 20}{3}=k

Now,

k=\dfrac{12-20}{3}\text{ and }k=\dfrac{12+20}{3}

k=\dfrac{-8}{3}\text{ and }k=\dfrac{32}{3}

Therefore, the two points are \left(0,\dfrac{-8}{3}\right)\text{ and }\left(0,\dfrac{32}{3}\right).

7 0
3 years ago
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