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morpeh [17]
2 years ago
15

Subtract 6 1/3 - 2 1/5

Mathematics
1 answer:
jolli1 [7]2 years ago
4 0

Answer:

4 2/15

Step-by-step explanation:

Convert both fractions to have the same denominator. Find the LCM of 3 and 5 and multiply by what you need to get there. 1x5=5, 3x5=15, so 1/3 becomes 5/15, and 1x3=3, 5x3=15, so 1/5 becomes 3/15. Finally, do normal subtraction, to get 4 2/15.

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When I solve this problem I get a big number, and I don't kwow if it's Ok:
Jlenok [28]

Answer:

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Step-by-step explanation:

6 0
3 years ago
I need some help please
gizmo_the_mogwai [7]

Answer:

a) -10

b) 7

Step-by-step explanation:

a) 2(x + 3) = x - 4

=  > 2x + 6 = x - 4

=  > 2x - x =  - 4 - 6

=  > x =  - 10

b) 4(5x - 2) = 2(9x + 3)

=  > 20x -8 = 18x + 6

=  > 20x - 18x = 8 + 6

=  > 2x = 14

=  > x =  \frac{14}{2}  = 7

3 0
2 years ago
Read 2 more answers
If you were to graph 1.14, 1.15, and 0.5. what would you count by on your number line?
GREYUIT [131]

If you want to include 0, the overall interval is 115 times 0.01, or 23 times 0.05 or 11.5 times 0.10. The latter might make it harder to plot 1.14, so I'd probably use an interval of 0.05.

Between 6 or 7 and about 25 intervals on a graph's scale are about right. More makes it pretty busy and sometimes difficult to tell which mark is associated with the number. A fewer number is indicated only if there are a fewer number of discrete values that need to be shown to adequately identify the data points.

6 0
3 years ago
Write the equation of the line that passes through (0, 3) and is parallel to the line 3x + 5y = 6.
Pavel [41]
Please, see the offered decision:
1) common equation for lines is y=kx+b. If k₁=k₂ (for line 1 and line 2) ⇒ 'line 1' || 'line 2'.
2) for line 3x+5y=6   k= -3/5. It means (according to item 1) for unknown line k is the same (-3/5).
3) using points (0;3) it is easy to find parameter b (x=0, y=3) via y=kx+b:
3=0*(-3/5)+b ⇔ b=3.
4) finaly (k=-3/5; b=3):
y=- \frac{3}{5}x+3 \ or \ 3x+5y=15
3 0
2 years ago
find the midpoint of AB. A(-9,3), B(7,-8). write an equation perpendicular to AB passes through the midpoint
steposvetlana [31]

Step-by-step explanation:

the solution is in the picture

3 0
1 year ago
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