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slava [35]
3 years ago
12

Solve -28<2x-10<-10

Mathematics
1 answer:
Sedbober [7]3 years ago
5 0
-28 < 2x - 10 < -10.....add 10 to all sections
-28 + 10 < 2x - 10 + 10 < -10 + 10....simplify
-18 < 2x < 0...divide all sections by 2
-18/2 < 2/2x < 0/2...simplify
-9 < x < 0 <===
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For this case we must solve the following equation:

x ^ 3 = 22

If we apply cubic root to both sides of the equation, then we eliminate the exponent on the left side:

\sqrt [3] {x ^ 3} = \sqrt [3] {22}\\x = \sqrt [3] {22}

That is the exact solution of the equation. Its decimal solution is given by:

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damaskus [11]

a. \frac{11}{12}

b. \frac{39}{20}

c. \frac{20}{21}

Step-by-step explanation:

Step 1; First, we convert the given fractions into improper ones. To do this, we multiply the whole number with the denominator of the fraction and add with it the same fraction's numerator whereas the denominator remains unchanged. To convert the fraction

3\frac{1}{4} = (3 × 4) + 1 / 4 = \frac{13}{4},

2\frac{1}{3} = (2 × 3) + 1 / 3 = \frac{7}{3},

6\frac{1}{5} = (6 × 5) + 1 / 5 = \frac{31}{5},

4\frac{1}{4} = (4 × 4) + 1 / 4 = \frac{17}{4},

5\frac{2}{7} = (5 × 7) + 2 / 7 = \frac{37}{7},

4\frac{1}{3} = (4 × 3) + 1 / 3 = \frac{13}{3}.

Step 2; Now we subtract, using LCM to arrive at the answer

3\frac{1}{4} - 2\frac{1}{3} = \frac{13}{4} - \frac{7}{3} = \frac{39-28}{12} = \frac{11}{12},

6\frac{1}{5} - 4\frac{1}{4} = \frac{31}{5} - \frac{17}{4} = \frac{124-85}{20} = \frac{39}{20},

5\frac{2}{7} - 4\frac{1}{3} = \frac{37}{7} - \frac{13}{3} = \frac{111-91}{21} = \frac{20}{21}.

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3 years ago
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