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Nookie1986 [14]
3 years ago
12

What is the answer to number 12

Mathematics
1 answer:
egoroff_w [7]3 years ago
5 0

how when there's no picture


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Perform the indicated operations; reduce the answer to lowest terms. a. 3⁄10 + 6⁄10 b. 1⁄3 + 1⁄4 + 1⁄6 c. 5⁄6 – 3⁄6 d. 2⁄3 – 6⁄1
Setler [38]

Answer:

a. 3⁄10 + 6⁄10 = 9/10

b. 1⁄3 + 1⁄4 + 1⁄6 = 3/4

c. 5⁄6 – 3⁄6 = 1/3

d. 2⁄3 – 6⁄10 = 1/15

e. 4⁄10 × 3⁄7 = 6/35

f. 1⁄6 × 6⁄15 = 1/15

g. 1⁄8 ÷ 4⁄9 = 9/32

h. 1⁄5 ÷ 3⁄4 = 4/15

Step-by-step explanation:

a. 3⁄10 + 6⁄10

= 3*1 + 6*1 / 10

= 3+6/10

= 9/10

b. 1⁄3 + 1⁄4 + 1⁄6

since denominators are different we take LCM of 3,4,6 which is 12

= 1*4 + 1*3 + 1*2 / 12

= 4+3+2/12

= 9 ÷ 3 / 12 ÷ 3

= 3 / 4

c. 5⁄6 – 3⁄6

= 5 - 3 / 6

= 2 ÷ 2 / 6 ÷ 2 = 1/3

d. 2⁄3 – 6⁄10

LCM of 3 and 10 is 30

= 2 * 10 - 6 * 3 / 30

= 20 - 18 / 30

= 2 ÷ 2 / 30 ÷ 2 = 1/15

e. 4⁄10 × 3⁄7

= 12 ÷ 2 / 70 ÷ 2 = 6/35

f. 1⁄6 × 6⁄15

= 6 ÷ 6/90 ÷ 6 = 1/15

g. 1⁄8 ÷ 4⁄9

= 1/ 8 * 9/4

=9/32

h. 1⁄5 ÷ 3⁄4

=1/5 * 4/3

= 4/15

3 0
3 years ago
Read 2 more answers
Select the correct answer.
MariettaO [177]

Answer:

you would use a demand chart

Step-by-step explanation:

5 0
3 years ago
Can someone help me with this Exponential Growth and Decay word problem? I have been able to get the rest of them done but this
Vinvika [58]

Answer:

A) 61.68

Step-by-step explanation:

This is an exponential decay of 0.875. The way to know that is to get that is\frac{right}{left} \\ (Pick a random number and divide by the number to the left of it) From there, there is a 12 hour difference from 7 am to 7pm! Divide 306.25/0.875 and divide it by 0.875 by the answer of that each time 12 times. The exact answer is 61.68402914

4 0
4 years ago
Read 2 more answers
The temperature at 4AM was-4C and at 7am 3C higher. What was the temperature at 5 am
trasher [3.6K]
It was -3C at 5 am because 7am-4am=3 and -4c=-3c=-1c so take of -1c to make it -3c.
7 0
3 years ago
Please help me this is the last one on my review
Anika [276]

try the very last one might be wrong but...

7 0
3 years ago
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