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Mrac [35]
1 year ago
6

Question 1 i need help on

Mathematics
1 answer:
ladessa [460]1 year ago
6 0

The actual distance between Bixby and Coyle is 105 miles which can be represented by 7 cm.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depends on other variable while a dependent variable is a variable that depends on other variable.

60 miles = 4 cm

Hence:

7 cm = 7 cm * (60 miles)/(4 cm)  = 105 miles

The actual distance between Bixby and Coyle is 105 miles which can be represented by 7 cm.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

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Is this what you needed?

4 0
2 years ago
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What is the answer to this math problem​
vesna_86 [32]

Hello from MrBillDoesMath!

Answer:

1/243

Discussion:  

3^(-3) *  3^(-2)                => as a^m * a^n = a^(m+n), the exponents add

= 3^ (-3 -2)

= 3^ (-5) =

= 1/ 3^5                           => as 3^5 = 243

= 1/243

which is approximately  .0041

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7 0
3 years ago
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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

5 0
3 years ago
Read 2 more answers
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