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Rom4ik [11]
2 years ago
10

How do you write out the domain and range of a function?

Mathematics
1 answer:
ra1l [238]2 years ago
6 0
X + 5 = 53 = \left[x \right] = \left[ 48\right][x]=[48]
you write out the domain and range of a function
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Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.
Rashid [163]

Answer:

The following pairs/results are matched:

  • 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33
  • 3\left(3t-4\right)-\left(2t+10\right) = 7t-22
  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}
  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Step-by-step explanation:

Lets solve all the expressions to match the results.

  • 5\left(2t+1\right)+\left(-7t+28\right)

<em>Solving the expression</em>

5\left(2t+1\right)+\left(-7t+28\right)

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

5\left(2t+1\right)-7t+28

10t+5-7t+28

3t+33

Therefore, 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33

  • 3\left(3t-4\right)-\left(2t+10\right)

<em>Solving the expression</em>

3\left(3t-4\right)-\left(2t+10\right)

9t-12-\left(2t+10\right)

9t-12-2t-10

7t-22

Therefore, 3\left(3t-4\right)-\left(2t+10\right) = 7t-22

  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)

<em>Solving the expression</em>

\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)

4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)

4t-\frac{8}{5}-3+\frac{4t}{3}

As

-3-\frac{8}{5}:\quad -\frac{23}{5}    and  \frac{4t}{3}+4t:\quad \frac{16t}{3}

So,

4t-\frac{8}{5}-3+\frac{4t}{3} will become \frac{16t}{3}-\frac{23}{5}

Therefore, \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}

  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)

<em>Solving the expression</em>

\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

-\frac{9}{2}t+3+\frac{7}{4}t+33

\mathrm{Group\:like\:terms}

\frac{9}{2}t+\frac{7}{4}t+3+33

\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t

-\frac{11}{4}t+3+33

-\frac{11}{4}t+36

Therefore, \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Thus, the following pairs/results are matched:

  • 5\left(2t+1\right)+\left(-7t+28\right) = 3t+33
  • 3\left(3t-4\right)-\left(2t+10\right) = 7t-22
  • \left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right) = \frac{16t}{3}-\frac{23}{5}
  • \left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right) = -\frac{11}{4}t+36

Keywords: algebraic expression

Learn more about algebraic expression from brainly.com/question/11336599

#learnwithBrainly

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Divide is by 100.

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Step-by-step explanation:

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156.73643

Step-by-step explanation:

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