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UkoKoshka [18]
3 years ago
9

Find the value of x.

Mathematics
1 answer:
Aneli [31]3 years ago
3 0

Answer:

39x

Step-by-step explanation:

You might be interested in
2x-8y=-24<br> 6x+4y=68<br> Using elimination method
nikklg [1K]
6(2x-8y=-24)
-2 (6x+4y=68)
12x-48y=-144
-12x-8y=-136
-56y=-280

-280÷-56=5
y=5

2x-8 (5)=-24
2x-40=-24
+40. +40
2x=16
16÷2=8
×=8


y=5
x=8
hope this helps, if needed a picture for better understanding,please comment.
4 0
3 years ago
Pls help I’m failing math I’ll brainlest
Keith_Richards [23]

Answer:

z=-9 - 5.4

hope it helps

Step-by-step explanation:

please mark me brainliest

4 0
3 years ago
I need help ASAP PLEASEE
Alex787 [66]

Answer:

9. 21

10. 70

11. 64

12. 31

13. 85

14. 31

double check the answer, i didnt use a calculator so might be wrong

4 0
3 years ago
Para embalar una caja se emplea 4,2 m de cinta adhesiva. ?Cuántas cajas se podrán embalar con tres rollos que tienen 3 hm, 7 dam
kompoz [17]

For this case, we perform the conversions:

First roll:

1 \ Hectometer --------> 100 \ meters\\3 \ Hectometer --------> x

x = \frac {3 * 100} {1}\\x = 300 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.

1 -----------> 4.2

c -----------> 300

c = \frac {300 * 1} {4.2}\\c = 71.42857143\\c = 71

You can pack 71 boxes.

Second roll:

1 \ Decametro --------> 10 \ meters\\7 \ Decameter --------> x\\x = \frac {7 * 10} {1}\\x = 70 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.

1 -----------> 4.2

c -----------> 70

c = \frac {70 * 1} {4.2}\\c = 16.66666667\\c = 16

You can pack 16 boxes.

Third roll:

1 -----------> 4.2

c -----------> 50

c = \frac {50 * 1} {4.2}\\c = 11.9047619\\c = 11

You can pack 11 boxes.

Thus, in total you can pack11 + 16 + 71 = 98 \ boxes

Answer:

98 boxes

4 0
3 years ago
PLEASE HELP ME
Anon25 [30]

equivalent to 5/24 is 15/72 (multiply top and bottom by 3)

answer is d. 15/72

7 0
3 years ago
Read 2 more answers
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