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alexgriva [62]
2 years ago
12

I ned help due in july 31st

Mathematics
1 answer:
Kamila [148]2 years ago
6 0

Answer:

71: 19    72: 100    73:9    74:1     75:-2      76:4

Step-by-step explanation:

71. 2(12)-5 = 24-5 = 19

72. 2(5(12)-10) = 2(60-10) = 2(50) =100

73. 3+12/2 = 3+6 = 9

74. 4(5)/12+4(2)= 20/20 = 1

75. 2(2)^2-10 = 8-10 = -2

76. 12-10+2 =4

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20 men decided to complete a work in 30 days. but after 6 days 8 men left now in how many day the work will be completed​
Serga [27]

First, we convert it into a unit like this

20 workers × 30 days = 600 units work

After leaving the 5 workers, we have 35 days to complete the work.

In 35 days only 15 workers will do the work.

So the work completed in 35 days by 15 workers = 35 ×15 =525 units

Remaining unit = 600 -525 =75 units

This unit would do by 5 workers in (75/5) = 15 days.

So after 15 days, 5 workers should leave the job.

6 0
3 years ago
3 questions pls help
3241004551 [841]

12. I need the graphs to answer it

11. y = 1/2x - 1

10. The first graph is the best

4 0
3 years ago
1)bir üçgenin kenar uzunlukları 2,3ve4 ile ters orantılıdır.üçgenin çevresi 39 cm olduğuna göre en uzun kenarı kaç cmdir?
goldenfox [79]
1. 
The edge lengths of a triangle are inversely proportional to 2,3 and 4.
The actual proportions of the lengths are then
12(1/2:1/3:1/4)=6:4:3
The longest side 
= 39* (6/(6+4+3)
= 39*6/13
=18

2. 63 breads : 35 kg flour
is the same as 
54 breads : x kg flour
Cross multiply and get
x=54*35/63=30
30 kg of flour are needed to make 54 breads.
3 0
3 years ago
5. Find the general solution to y'''-y''+4y'-4y = 0
CaHeK987 [17]

For any equation,

a_ny^(n)+\dots+a_1y'+a_0y=0

assume solution of a form, e^{yt}

Which leads to,

(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0

Simplify to,

e^{yt}(y^3-y^2+4y-4)=0

Then find solutions,

\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}

For non repeated real root y, we have a form of,

y_1=c_1e^t

Following up,

For two non repeated complex roots y_2\neq y_3 where,

y_2=a+bi

and,

y_3=a-bi

the general solution has a form of,

y=e^{at}(c_2\cos(bt)+c_3\sin(bt))

Or in this case,

y=e^0(c_2\cos(2t)+c_3\sin(2t))

Now we just refine and get,

\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}

Hope this helps.

r3t40

5 0
3 years ago
2.(03.01 LC)
strojnjashka [21]
Answer:
7(3 + 7)
Why:
7x3=21
7x7=49
4 0
3 years ago
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