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Mars2501 [29]
3 years ago
11

What is 0.46 rounded to the nearest tenth

Mathematics
2 answers:
andrey2020 [161]3 years ago
7 0

Answer: 0.5

Step-by-step explanation: Use this for rounding. 5 or more, raise the score. 4 or less, let it rest.

matrenka [14]3 years ago
5 0

Answer:

.50

Step-by-step explanation:

if the hundredths place is higher than 4, you round up.

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Is this graph proportional
Zanzabum
Yes the graph is proportional
5 0
3 years ago
10. MJ and Peter Parker leave school traveling in opposite
Paladinen [302]

Using proportions, it is found that:

  • Peter is walking at a rate of 3 km/h.
  • MJ is biking at a rate of 9 km/h.

---------------------------

  • Peter is walking at a rate of x km/h.
  • MJ is biking 6 km/h faster than Peter, thus 6 + x km/h.
  • Opposite directions, thus, their distance increases at a rate of (6 + 2x) km each hour, as 6 + x + x = 6 + 2x.

6 + 2x km each hour, 18 km after 1.5 hours, thus, the rule of three is:

1h - (6 + 2x) km

1.5h - 18 km

Applying cross multiplication:

1.5(6 + 2x) = 18

9 + 3x = 18

3x = 9

x = \frac{9}{3}

x = 3

x + 6 = 3 + 6 = 9

Then:

  • Peter is walking at a rate of 3 km/h.
  • MJ is biking at a rate of 9 km/h.

A similar problem is given at brainly.com/question/24112433

6 0
2 years ago
10.
alukav5142 [94]

Step-by-step explanation:

ang hirap naman ng pinasasagot mo

8 0
2 years ago
Solve 2x+y=3 x+y=5 in substitution method
Nastasia [14]
2x+y=3
- x+y=5
---------------
x=-2

2(-2)+y=3
y=7

-2+y=5
y=7

(-2,7)
4 0
3 years ago
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

5 0
1 year ago
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