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faltersainse [42]
3 years ago
14

What is equivalent to (x+1)^2

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
6 0
What is x+1^2 is equivalent
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What is the slope of a line that is perpendicular to the graphed line?
Lemur [1.5K]

Answer:

2

Step-by-step explanation:

To find the slope of the graphed line, you take two points. I have chosen (0, 4) and (8, 0)

The formula of a slope is (y value of second point -y value of first point) divided by (x value of second point -x value of first point):

\frac{0-4}{8-0} =-\frac{1}{2}

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

The slope for a perpendicular line is 2.

6 0
3 years ago
Hey! Can somebody please help my friend with 5,211 ÷ 6? I need to have a picture of the work if somebody can please do that!
Marysya12 [62]

Answer:

868.5

Step-by-step explanation: If u devide by groups, you can find the answer.

6 0
3 years ago
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Jacob took out a loan for $950 at a 13.2% APR, compounded monthly, to buy
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Answer:

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Step-by-step explanation:

just answered it

3 1
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
Are the answers for 1-3 correct? I need help with 4 and 5!!! Please!
vovangra [49]
1.

a. 12d² -6d

b. 6c^5 + 8c^4 -10c³

c. correct

2. 

a. correct

b. correct

c. 12r^8-6r^4 + 9r^2, then multiply the rest by -1

3. correct

4. x² + 10

5. d=4, -1/3
8 0
3 years ago
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