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  1. Home
  2. 12th grade
  3. Precalculus
  4. Exercise : Evaluate an expression that involves any operation

Evaluate an expression that involves any operation Precalculus

Evaluate \dfrac{3\times a + b}{c} with a=2.3, b=\dfrac25, c=\dfrac45

First, compute the numerator of this fraction. Notice that:

b = \dfrac{2}{5}= 0.4

Therefore :

3 \times a + b = 3 \times 2.3 + 0.4 = 6.9 + 0.4 = 7.3 = \dfrac{73}{10}

We have:

\dfrac{3 \times a + b}{c} = \left(3 \times a + b\right) \times \dfrac{1}{c} =\dfrac{73}{10} \times \dfrac{5}{4}=\dfrac{73}{8}

\dfrac{3\times a + b}{c}=\dfrac{73}{8}

Evaluate \left(\dfrac{3}{5}\right) \times \left(m + \dfrac{n^2}{4}\right) with m = 1.3 and n = 2

We have:

m + \dfrac{n^2}{4} = 1.3 + \dfrac{2^2}{4}= 1.3 + \dfrac{4}{4} = 1.3 + 1 = 2.3

Note that:

2.3 = \dfrac{23}{10}

Therefore:

\dfrac{3}{5} \times 2.3 = \dfrac{3}{5} \times \dfrac{23}{10} = \dfrac{69}{50}

\left(\dfrac{3}{5}\right) \times \left(m + \dfrac{n^2}{4}\right) = \dfrac{69}{50}

Evaluate \dfrac{\left(3 - \left(-4\right)x\right)^2}{y + 15} with x = 2 and y=7

First, compute the numerator:

\left(3 - \left(-4\right)x\right)^2 = \left(3 + 4x\right)^2 = \left(3 + 4 \times 2\right)^2 = \left(3 + 8 \right)^2 = 11^2 = 121

In the denominator:

y+15=7+15=22

Thus:

\dfrac{\left(3 - \left(-4\right)x\right)^2}{y+15} = \dfrac{121}{22}

Simplify:

\dfrac{\left(3 - \left(-4\right)x\right)^2}{y + 15} = \dfrac{11}{2}

Evaluate \dfrac{-4 \times v}{c^2} + 2.6 with v = -2 and c = 4

-4 \times v = -4 \times -2 = 8

c^2 = 4^2 = 16

Thus:

\dfrac{-4 \times v}{c^2} = \dfrac{8}{16}=\dfrac{1}{2} = 0.5

Therefore:

0.5 + 2.6 = 3.1

\dfrac{-4 \times v}{c^2} + 2.6 =3.1

Evaluate \dfrac{6 \times s - t}{t + w} with s = -1, t = 2, and w = .75

First, compute the numerator:

6 \times s - t = 6\times -1 - 2 = -6 - 2 = -8

In the denominator:

t + w = 2 + 0.75 = 2.75

Note that:

2.75 = \dfrac{11}{4}

Therefore:

\dfrac{-8}{11/4} = -8 \times \dfrac{4}{11} = \dfrac{-32}{11}

\dfrac{6 \times s - t}{t + w} = \dfrac{-32}{11}

Evaluate \dfrac{\left(5 - a\right)^2\times 4}{b + \dfrac{2}{3}} with a = 5.5 and b = \dfrac{1}{6}

First, compute the numerator:

\left(5 - a\right)^2 = \left(5 - 5.5\right)^2= \left(-0.5\right)^2 = 0.25

Note that:

0.25 \times 4 = 1

In the denominator:

b + \dfrac{2}{3} = \dfrac{1}{6} + \dfrac{2}{3} = \dfrac{1}{6} + \dfrac{4}{6} = \dfrac{5}{6}

Therefore:

\dfrac{\left(5 - a\right)^2 \times 4}{b + \dfrac{2}{3}} = \dfrac{1}{5/6}= 1 \times \dfrac{6}{5} = \dfrac{6}{5}

\dfrac{\left(5 - a\right)^2\times 4}{b + \dfrac{2}{3}} = \dfrac{6}{5}

Evaluate \dfrac{2 - a \times b}{\left(-3\right)^2 - c + 0.5} with a = -2, b = 0.75, and c = 2.5

First, compute the numerator:

2 - a \times b = 2 - \left(-2\right)\times 0.75 = 2 - \left(-1.5\right) = 3.5

In the denominator:

9 - c + 0.5 = 9 - 2.5 + 0.5 = 6.5 + 0.5 = 7

Therefore:

\dfrac{2 - a \times b}{9 - c + 0.5} = \dfrac{3.5}{7}= \dfrac{1}{2} \text{ or } 0.5

\dfrac{2 - a \times b}{9 - c + 0.5}= 0.5

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See also
  • Course : Fundamentals of numbers
  • Exercise : Convert a sentence into a mathematical expression or an equation
  • Exercise : Classify real numbers
  • Exercise : Convert from decimals to fraction
  • Exercise : Multiply real numbers
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