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lyudmila [28]
3 years ago
12

Write the solution to the given inequality in interval notation.

Mathematics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

The answer should be C.

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6a+4=61/3+4<br> I don’t know what I’m doing wrong because I keep getting 10 1/3
Vlad1618 [11]

Answer:

a = 3 7/18 or 3.388889

Step-by-step explanation:

First, you want to get the a by itself, so you would subtract 4 from the left side of the equals sign.

6a + 4 = 61 / 3 + 4

    -4                   -4

6a = 61 / 3

By doing that, they cancel each other out, so you are left with

6a = 61 / 3

Then you want to get a by itself, so you divide 6 from both sides.

6a / 6 = 61 / 3  /  6

Remember that 6 can also be written as 6/1

Remember that dividing a fraction is multiplying the reciprocal.

6a / 6 = 61 / 3  /  6 / 1

When dividing fractions remember to Skip the first fraction, Flip the second fraction, and Multiply the two fractions together.

61 / 3      x        1 / 6

skip   multiply   flip

Multiply across the top, and across the bottom.

61(1) and 3(6)

Then you get

61 / 18

So a = 61 / 18

If you want to convert that to a mixed number, it would be 3  7 / 18

If you want that in a decimal, it would be 3.388889

4 0
2 years ago
You buy a 24 pack of 591mL bottles of water. How many Liters is that?
4vir4ik [10]
24 bottles x 591 mL/bottle = 14,184 mL

Since 1L = 1000 mL,

14,184 mL x 1L/1000mL = 14,184/1000 = 14.184L (14 whole liters)

3 0
3 years ago
8xy^3*xy ^8 plsssssssssssssssssssss help
Sindrei [870]

Answer:

x • (x^3 - 3x^2y + 3xy^2 + 7y^3)

Step-by-step explanation:

(8x • (y^3)) +  x • (x - y)^3

2^3xy^3 +  x • (x - y)^3

Evaluate :  (x-y)^3   =   x^3-3x^2y+3xy^2-y^3

Pull out like factors :

x^4 - 3x^3y + 3x^2y^2 + 7xy^3  =

x • (x^3 - 3x^2y + 3xy^2 + 7y^3)

x^3 - 3x^2y + 3xy^2 + 7y^3 is not a perfect cube

4 0
2 years ago
No files or links please give me answer
alexgriva [62]

Answer:

I think it is J if not it may be H if they are both wrong i am so so sorry

Step-by-step explanation:

6 0
2 years ago
Why is 11,12,12 a Pythagorean triple
DENIUS [597]
11, 12, 12 is not a triple
8 0
3 years ago
Read 2 more answers
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