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jeka57 [31]
3 years ago
13

35 over 11 rounded to the nearest tenth

Mathematics
2 answers:
adoni [48]3 years ago
6 0

Answer:

im gonna say 40

Step-by-step explanation:

(if i got it wrong im so so sorry

olga55 [171]3 years ago
5 0
35 over 11 rounded to the nearest tenth would be 3.2
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I’m just confused help!
rusak2 [61]

Answer:

Step-by-step explanation:

180

4 0
3 years ago
Translate this phrase into an algebraic expression.
Roman55 [17]

Answer:

8+2j like you say let j be the variable and more than means+

7 0
3 years ago
Read 2 more answers
-6x-14y=16 and -2x+7y=17 solve by elimination show the work
professor190 [17]

Answer:

(- 5, 1 )

Step-by-step explanation:

- 6x - 14y = 16 → (1)

- 2x + 7y = 17 → (2)

Multiplying (2) by - 3 and adding to (1) will eliminate the x- term

6x - 21y = - 51 → (3)

Add (1) and (3) term by term to eliminate x

0 - 35y = - 35

- 35y = - 35 ( divide both sides by - 35 )

y = 1

Substitute y = 1 into either of the 2 equations and solve for x

Substituting into (1)

- 6x - 14(1) = 16

- 6x - 14 = 16 ( add 14 to both sides )

- 6x = 30 ( divide both sides by - 6 )

x = - 5

solution is (- 5, 1 )

6 0
2 years ago
Match each equation with its solution set.
taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

 magnitude of x is always positive

Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I  I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

∴ x = 9

OR

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I  I never give negative answer

* The solution of Ix - 3I = -12 is ∅

4 0
3 years ago
Given {27,15,3,-9,...} find a11
coldgirl [10]

Answer: -21, -33, -45

Step-by-step explanation:

each next number is being subtracted by 12 so

4 0
4 years ago
Read 2 more answers
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