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ZanzabumX [31]
3 years ago
13

Use substitution to determine which of the following points is a solution to the standard form equation below. 2y = 5

Mathematics
1 answer:
sashaice [31]3 years ago
7 0
The question is incomplete.

This is the complete question as I found in internet:

<span>Use substitution to determine which of the following points is a solution to the standard form equation below 5x-2y = 10

these are the points:
</span>
-1,5

1,5

0,-5

0,5

Answer: (0, -5)

Explanation:


point          x           y        5x - 2y = 10 ?


-1,5         -1           5         5(-1) - 2(5) = - 5 - 10 = - 15 ≠ 10 ⇒ not a solution

1,5            1          5         5(1) - 2(5) = 5 - 10 = 5 ≠ 10 ⇒ not a solution
 
0,-5           0        -5           0 -2(-5) = 10 ⇒ a solution

0,5             0         5          0 - 2(5) = - 10 ≠ 10 ⇒ not a solution

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What are the types of roots of the equation below?<br> - 81=0
Tju [1.3M]

Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0. This can be obtained by finding root of the equation using algebraic identity.    

<h3>What are the types of roots of the equation below?</h3>

Here in the question it is given that,

  • the equation x⁴ - 81 = 0

By using algebraic identity, (a + b)(a - b) = a² - b², we get,  

⇒ x⁴ - 81 = 0                      

⇒ (x² +  9)(x² - 9) = 0

⇒ (x² + 9)(x² - 9) = 0

  1. (x² -  9) = (x² - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
  2. x² + 9 = 0 ⇒ x² = -9 ⇒ x = √-9 ⇒ x= √-1√9 ⇒x = ± 3i

⇒ (x² + 9) = (x - 3i)(x + 3i)

Now the equation becomes,

[(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

Therefore x + 3, x - 3, x + 3i and x - 3i are the roots of the equation

To check whether the roots are correct multiply the roots with each other,

⇒ [(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

⇒ [x² - 3x + 3x - 9][x² - 3xi + 3xi - 9i²] = 0

⇒ (x² +0x - 9)(x² +0xi - 9(- 1)) = 0

⇒ (x² - 9)(x² + 9) = 0

⇒ x⁴ - 9x² + 9x² - 81 = 0

⇒ x⁴ - 81 = 0

Hence Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: What are the types of roots of the equation below?

x⁴ - 81 = 0

A) Four Complex

B) Two Complex and Two Real

C) Four Real

Learn more about roots of equation here:

brainly.com/question/26926523

#SPJ9

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∠1 and ∠2 are complementary. If m∠1 = 4m∠2, find m∠1
77julia77 [94]

Answer:

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Step-by-step explanation:

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Hence ∠1 = 90 - 18 = 72°

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