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Llana [10]
3 years ago
9

Solve this because I'm horrible at maths

Mathematics
2 answers:
m_a_m_a [10]3 years ago
5 0

\tt \sqrt{1989 \times 1989 + 1989 \times 1989}  + 13 - 2007

  • Write the repeated multiplication in exponential form

\tt \sqrt{   {1989}^{2} +  {1989}^{2}  }  + 13 - 2007

  • Collect like terms

\tt \sqrt{2 \times  {1989}^{2} }   + 13 - 2007

Simplify the radical expression

\tt1989 \sqrt{2}  + 13 - 2007

Calculate the difference

\tt1989 \sqrt{2}  - 1994

Calculate the approx value

\tt \approx818.87

scoray [572]3 years ago
3 0

Answer:

818.87

explanation:

\hookrightarrow \sf \sqrt{1989\cdot 1989+1989\cdot 1989}+13-2007

\hookrightarrow \sf \sqrt{1989\cdot \:1989+1989\cdot \:1989}-1994

\hookrightarrow \sf 1989\sqrt{2}-1994

\hookrightarrow \sf 818.8707756

\hookrightarrow \sf 818.87

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Sharon earns $25 per item she sales plus a base salary of $100 per week. Write and solve an inequality to find how many items sh
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4 years ago
How do I solve this word problem?
german

Answer:

  • 280 student tickets
  • 520 adult tickets

Step-by-step explanation:

You may recognize that you are given two relationships between two unknowns. You can write equations for that.

You are asked for numbers of adult tickets and of student tickets. It often works well to let the values you're asked for be represented by variables. We can choose "a" for the number of adult tickets, and "s" for the number of student tickets. Then the problem statement tells us the relationships ...

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(You are supposed to know that the revenue from selling "a" adult tickets is found by multiplying the ticket price by the number of tickets: 12.50a.)

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You can solve these two equations any number of ways. One way is to do it by <em>elimination</em>. We can multiply the first equation by 12.50 and subtract the second equation:

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  5s = 1400 . . . . simplify. (The "a" variable has been eliminated.)

  s = 280 . . . . . . divide by 5

Then the number of adult tickets can be found from the first equation:

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4 years ago
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