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bulgar [2K]
3 years ago
15

Find the missing side of each triangle. Leave your answers in simplest radical form.

Mathematics
2 answers:
rosijanka [135]3 years ago
4 0

Answer:

Step-by-step explanation:

pythagoras theorem

a^2+b^2=c^2

8^2+9^2=x^2

64+81=x^2

145=x^2

\sqrt{145=x

DENIUS [597]3 years ago
3 0

Answer:

The missing side of the triangle is 12.04 mi / √145 mi.

Step-by-step explanation:

<h3>Given :- </h3>
  • Perpendicular = 9 mi
  • Base = 8 mi
  • And Hypotenuse = missing side = x

<h3>Solution :- </h3>

Now substitute the values

➟ ( 9 mi ) ² + ( 8 mi ) ² = x²

Simplify the values

➟ 81 mi ² + 64 mi ² = x ²

Add the values

➟ 145 mi ² = x ²

Finding , the root of 145 mi ²

➟ √145 mi ² = x

➟ 12.04 mi = x

Therefore, The missing side of the triangle is 12.04 mi / √145 mi.

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Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.
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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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