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maks197457 [2]
3 years ago
6

does anyone have a brainly plus account I can borrow? I have a lot of classes and I don't have access to a debit/credit card.​

Mathematics
2 answers:
Mariulka [41]3 years ago
5 0

Step-by-step explanation:

Im in the same boat. If you Uninstall and reinstall it you get more of the thingys to ask questions. (I learned that today)

posledela3 years ago
3 0

Try uninstalling and then reinstalling

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Solve the following inequality for X <br> -7.3x &lt; -21.9x + 73
Aleonysh [2.5K]

Answer:

X=®™✓©™[[

Step-by-step explanation:

¯\_༼ ಥ ‿ ಥ ༽_/¯

7 0
2 years ago
If AB=48 and BC=22, then AC=?
alexandr1967 [171]

If A, B and C are collinear, then

1) if B is between A and C:

AC = AB + BC

AC = 48 + 22 = 70

2) if C is between A and B:

AB = AC + BC

48 = AC + 22 |-22

AC = 26

3) if A is between B and C:

BC = AB + AC

22 = 48 + AC |-48

AC = - 26 < 0 FALSE

Answer:

if B is between A and C, then AC = 70

if C is between A and B, then AC = 26

4 0
3 years ago
Read 2 more answers
Decompose the fraction (3/4) in at least 5 ways?
gulaghasi [49]

Answer:

Step-by-step explanation:

3/4

6/8

12/16

24/ 32

48/ 64

9/12

4 0
3 years ago
5 quarters weigh approximately 1 ounce. How many ounceswould you expect a collection of quarters worth $15 to weigh?
PSYCHO15rus [73]

5 quarters =1.25$

15$÷1.25$=12

then 15$ in quaters would weigh 12 ounces

8 0
2 years ago
Given the Arithmetic series A1+A2+A3+A4 7 + 11 + 15 + 19 + . . . + 91 What is the value of sum?
aivan3 [116]

Answer:

The value of the sum is 1078

Step-by-step explanation:

* Lets revise how to find the sum of the arithmetic series

- In the arithmetic series there is a constant difference between each

 two consecutive terms

- Ex:

# 4 , 9 , 14 , 19 , 24 , .......... (+ 5)

# 25 , 15 , 5 , -5 , -15 , .......... (-10)

- So if the first term is a and the constant difference between each two

  consecutive terms is d, then

  U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , ..........

- Then the nth term is Un = a + (n - 1)d

- The sum of n terms is Sn = n/2[a + L] , where L is the last term in

 the series

* Lets solve the problem

∵ The arithmetic series is 7 + 11 + 15 + 19 + ......................... + 91

∴ The first term (a) is 7

∴ The last term is (L) 91

- Lets find the constant difference

∵ 11 - 7 = 4 and 15 - 11 = 4

∴ The constant difference (d) is 4

- Lets find the sum of the series from 7 to 91

∵ Sn = n/2[a + L]

∵ a = 7 and L = 91

∴ Sn = n/2[7 + 91]

- We need to know how many terms in the series

∵ L is the last term and equals 91 lets find its position in the series

∵ Un = a + (n - 1)d

∵ a = 7 , d = 4 and Un = 91

∴ 91 = 7 + (n - 1)(4) ⇒ subtract 7 from both sides

∴ 84 = (n - 1)(4) ⇒ divide both sides by 4

∴ 21 = n - 1 ⇒ add 1 to both sides

∴ n = 22

∴ The number of the terms in the series is 22

- Lets find the sum of the 22 terms (S22)

∴ S22 = 22/2[7 + 91]

∴S22 = 11[98] = 1078

* The value of the sum is 1078

4 0
3 years ago
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