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Mrrafil [7]
2 years ago
12

4x+2>10 what is the answer to this?

Mathematics
2 answers:
zhenek [66]2 years ago
6 0

Answer:

x > 2

Step-by-step explanation:

4x + 2 > 10 \\  \\ 4x > 10 - 2 \\  \\ 4x > 8 \\  \\ x >  \frac{8}{4}  \\  \\ x > 2

Olenka [21]2 years ago
5 0

Answer:

x > 2

Step-by-step explanation:

4x + 2 > 10

     - 2   - 2

subtract two from each side

4x > 8

then divide 4 from each side

x > 2

Hope this helps dude

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Plz help with this word problem
Andreyy89
He will mix some 20% solution and some 70% solution to make 60% solution.

Let x = number of liters of 20% solution.
Let y = number of liters of 70% solution.

He wants to make 50 liters of 60% solution, so

x + y = 50   First Equation

The amount of acid in x amount of 20% solution is 20% of x, or 0.2x
The amount of acid in y amount of 70% solution is 70% of y, or 0.7y
The amount of acid in 50 liters of 60% solution is 60% of 50 liters, or 0.6 * 50 = 30
Now we add the amounts of acid.
0.2x + 0.7y = 30   Second Equation

x + y = 50
0.2x + 0.7y = 30

-0.2x - 0.2y = -10
0.2x + 0.7y = 30
------------------------
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y = 40

x + y = 50

x + 40 = 50

x = 10

Answer: He needs 10 liters of 20% solution and 40 liters of 70% solution.
4 0
3 years ago
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FAST!! Evaluate tan60/cos45<br> √6<br> √3/2<br> √2/3<br> 1√6
evablogger [386]

Answer:

\frac{\tan 60\degree}{\cos45 \degree}= \sqrt{6}

Step-by-step explanation:

We want to evaluate

\frac{\tan 60\degree}{\cos45 \degree}

We use special angles or the unit circle to obtain;

\frac{\tan 60\degree}{\cos45 \degree}=\frac{\sqrt{3}}{\frac{\sqrt{2}}{2}}

This implies that;

\frac{\tan 60\degree}{\cos45 \degree}=\sqrt{3}\div \frac{\sqrt{2}}{2}

\frac{\tan 60\degree}{\cos45 \degree}=\sqrt{3}\times \sqrt{2}

\frac{\tan 60\degree}{\cos45 \degree}= \sqrt{6}

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2 years ago
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erastovalidia [21]

Answer:

B

Step-by-step explanation:

16 x 3/2 = 24

24 x 3/2 = 36

36 x 3/2 = 54

8 0
3 years ago
What is the answer to -25+5=
lawyer [7]
Answer to the question

-20

7 0
2 years ago
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lawyer [7]
Im pritty sure it would be 30%

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3 years ago
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