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Genrish500 [490]
3 years ago
8

Subtract. Write your answer in simplest form. 7 1/15 - 3 4/5

Mathematics
2 answers:
Mnenie [13.5K]3 years ago
7 0

Answer:

3\frac{4}{15}

Step-by-step explanation:

7\frac{1}{15}

7=6+1

=6+1\frac{1}{15}

a\frac{b}{c}=\frac{a\cdot \:c+b}{c}

1\frac{1}{15}=\frac{1\cdot 15+1}{15}=\frac{16}{15}

=6+\frac{16}{15}

=6\frac{16}{15}

=6\frac{16}{15}-3\frac{4}{5}\\

\mathrm{Subtract\:the\:numbers:}\:6-3=3

3

\mathrm{Combine\:fractions}\:\frac{16}{15}-\frac{4}{5}:\quad \frac{4}{15}

\frac{16}{15}-\frac{4}{5}

\mathrm{Least\:Common\:Multiplier\:of\:}15,\:5:\quad 15

=\frac{16}{15}-\frac{12}{15}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

=\frac{16-12}{15}

\mathrm{Subtract\:the\:numbers:}\:16-12=4

=\frac{4}{15}\\

=3+\frac{4}{15}\\

=3\frac{4}{15}

Bingel [31]3 years ago
5 0

Answer:

3 4/15

Step-by-step explanation:

= 7 + 1 /15 - 3 -4/5

7-3 =4

4/5 = 12/15

1/15-12/15

3 ⁴/⁵

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Find the amount of each payment to be made into a sinking fund which earns 5â% compounded quarterly and produces â$55 comma 000
ankoles [38]

Answer:

$2743.66

Step-by-step explanation:

Data provided in the question:

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Time, t = 4.5 years

Now,

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Interest rate per period = 5% ÷ 4 = 1.25% = 0.0125

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7 0
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user100 [1]

Answer:

\frac{30x+30}{x+7} is the simplified radical expression.

Step-by-step explanation:

For this one, we have fractions in the bottom of a fraction (denominator).

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\frac{5x+5}{\frac{1}{6}x+\frac{7}{6}  }

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\frac{30x+30}{x+7} is the simplified radical expression.

4 0
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Use the matrix tool to solve the system of equations. Choose the correct ordered pair.3x + 5y = 42 6x + 5y = 54
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Solutions  

In Matrix we use initially based on systems of linear equations.The matrix method is similar to the method of Elimination as but is a lot cleaner than the elimination method.Solving systems of equations by Matrix Method involves expressing the system of equations in form of a matrix and then reducing that matrix into what is known as Row Echelon Form.<span> 

Calculations 

</span>⇒ <span>Rewrite the linear equations above as a matrix 
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\left[\begin{array}{ccc}3&5&42\\6&5&54\\\end{array}\right] 

⇒  Apply to Row₂ : Row₂ - 2 <span>Row₁ 
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\left[\begin{array}{ccc}3&5&42\\0&-5&-30\\\end{array}\right] 

⇒ <span>Simplify rows
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\left[\begin{array}{ccc}3&5&42\\0&1&6\\\end{array}\right] 

Note: The matrix is now in echelon form.
<span>The steps below are for back substitution. 
</span>
⇒ Apply to Row₁<span> : Row</span>₁<span> - </span>5 Row₂ 

\left[\begin{array}{ccc}3&0&12\\0&1&6\\\end{array}\right] 

⇒ <span>Simplify rows 
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\left[\begin{array}{ccc}1&0&4\\0&1&6\\\end{array}\right] 

⇒ <span>Therefore, 
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x=4&#10;&#10;y = 6
<span> 

 
</span>

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Answer:

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Step-by-step explanation:

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  P = 2(l +w) = 2(4 +3) = 2(7)

  P = 14 . . . . units

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