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den301095 [7]
3 years ago
13

The solution to 5.2x = 14.04 is x = ___.

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
7 0
The frist is 2.7
the 2nd is 2.34
the 4th is 8
the 5th is 4
I think the 3rd is 11over8
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Is that the only ones I used
jek_recluse [69]

Answer:

yes

Step-by-step explanation:

7 0
3 years ago
Y = 5+5(x-8)<br> Can someone please solve this?
vekshin1

y = 5+5(x-8)

y = 5 + 5x - 40

y = 5x - 35

7 0
4 years ago
how many liters of a 40% acid solution must be added to 12 liters of a 20% solution to obtain a 25% solution​
kkurt [141]

Answer:

The answer is 4 liters.

Step-by-step explanation:

We want a final 25% solution, which means that the solvent should constitute 25 percent of all the liters of solution.

Let us call x the litres of 40% solution, so when they are added to the 12 litres of 20% solution, the total litres of the new solution are:

x+12

of these liters, 25% must be the solvent:

N=0.25(x+12)

Now for the 40% solution, the liters of the solvent are 0.4x, and for the 20% liter solution, the liters of solve are 0.2*12=2.4 liters. Thus the total liters N of the solvent are

N=0.4x+2.4

Therefore

N=0.4x+2.4=0.25(x+12)

\therefore x=4\:litres

6 0
3 years ago
3(m + 5 + 9m)<br><br> Part A: Write two expressions that are equivalent to the given expression.
mihalych1998 [28]
3 (m + 5 + 9m)

Two expressions that are e<span>quivalent to the given expression are:

3m + 15 + 27m 

or 

30m + 15 

you get both expressions by multiplying the original expression 

the second expression is obtained by adding the like terms</span>
5 0
4 years ago
Read 2 more answers
The response to a question has three alternatives: A, B, and C. A sample of 120 responses provides 60 A, 12 B, and 48 C. Show th
zimovet [89]

Answer:

CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

Step-by-step explanation:

Given that;

the frequencies of there alternatives are;

Frequency A = 60

Frequency B = 12

Frequency C = 48

Total = 60 + 12 + 48 = 120

Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;

Relative Frequency A = Frequency A / total = 60 / 120 = 0.5

Relative Frequency B = Frequency B / total = 12 / 120 = 0.1

Relative Frequency C = Frequency C / total = 48 / 120 = 0.4

therefore;

CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

5 0
3 years ago
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