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sasho [114]
3 years ago
12

What is another way to write x 4/3

Mathematics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

1.3333333333333333 or 1 1/3

Step-by-step explanation:

What is the x?

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Write an equation to find x by substituting.
Harlamova29_29 [7]

Answer:

3

Step-by-step explanation:

5x-2=3x+4

5x-3x=4+2

2x=6

x=3

5 0
3 years ago
If a student wanted to immediately eliminate (get rid of) the fraction in this problem, they should:
mash [69]

Answer:

2x-15=12

Step-by-step explanation:

4 0
3 years ago
Will mark brainliest!!!!!<br> complete this statement<br> 15x^4a^2+20x^2a^3=5x^2a^2 ( )
Angelina_Jolie [31]
15x^4a^2+20x^2a^3=5x^2a^2 (3x^2+ 4a )

Answer: 3x² + 4a
5 0
3 years ago
Solve the equation
inysia [295]

Answer:

x ≈ 1.32, x ≈ - 5.32

Step-by-step explanation:

Given

x² + 4x - 7 = 0 ( add 7 to both sides )

x² + 4x = 7

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(2)x + 4 = 7 + 4

(x + 2)² = 11 ( take the square root of both sides )

x + 2 = ± \sqrt11} ( subtract 2 from both sides )

x = - 2 ± \sqrt{11}

Thus

x = - 2 - \sqrt{11} ≈ - 5.32 ( to 2 dec. places )

x = - 2 + \sqrt{11} ≈ 1.32 ( to 2 dec. places )

8 0
3 years ago
Read 2 more answers
11. Write the balanced equation for the reaction of HC2H3O2, with Al(OH)3 to form H2O and Al(C2H302)3:
Kamila [148]

Answer:

74.36 g of aluminium acetate.

730.27g of aluminium acetate.

- to the nearest hundredth.

Step-by-step explanation:

Acetic acid is usually written as CH3COOH.

a. 6CH3COOH + Al(OH)3 ---> Al(CH3COO)3 + 9H2O

So 6 moles of acetic acid produce 1 mole of aluminium acetate.

Using the molecular masses

6*( 1.008*4 + 12.011*2 + 16 *2) g acetic acid gives (26.98+3(36.032+ 2*12.011)

348.228 g acetic acid gives 207.142 g Al acetate.

So 125 g gives (207.142 / 348.228) * 125

= 74.36 g of aluminium acetate.

b.

(26.98 + 3*16 + 3 * 1.008) g  of Al(OH)3 gives 207.142 g Al acetate

78.004 g gives 207.142 g Al acetate

275 g gives   (207.142 / 78.004) * 275

= 730.27g Al acetate.

7 0
3 years ago
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