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3 years ago

What are the domains and ranges of the graphs

Mathematics

1 answer:

Maslowich3 years ago

7 0

Answer:

The domain for graph 1 is all real numbers, the range is y >= 0.

The domain for graph 2 is x >= 0, the range is y >= 0

Step-by-step explanation:

Graph 1: The x values are infinite for the graph, the y values will always be above zero and continue to be infinite.

Graph 2: The x values start at 0 and go to the right for infinity, the y values start at 0 and continue to infinity.

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Can some one help quickly. Please show work

Sphinxa [80]

Answer: the answer is 2

Step-by-step explanation:

6 0

3 years ago

Determine the angular velocity, in radians per second, of 14.2 revolutions in 8 seconds

adoni [48]

To answer this item, we use dimensional analysis and conversion factor, 1 revolution is equal to 2π rad. Using the concept and the given,

n = (14.2 revolutions / 8 seconds)(2π rad/1 rev)
n = 11.15 rad/s
Thus, the angular velocity is equal to 11.15 rad/s.

8 0

3 years ago

the sum of the first 6 terms of a geometric series is 15624 and the common ratio is 5. what is the first term of the series?

9966 [12]

Answer:

first term a=4

Step-by-step explanation:

Sum of the first 6 terms S6 = 15624

common ratio r= 5

n= 6

first term a=?

Formula: S6= a(r^n-1)/r-1

15624 = a( 5^6-1)/5-1

15624 = a(15625-1)/4

15624= a( 15624)÷4

15624= 15624a÷4

Cross multiply

15624×4 = 15624a

62496 = 15624a

divide both sides by 15624

62496÷15624 = 15624a÷15624

4 = a

a= 4

7 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 - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
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

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5 0

1 year ago

How are the perimeters of the preimage polygon and the image polygon related? Explain.

Romashka-Z-Leto [24]

The difference between Pre-Image and Image is given as follows:

Image is the shape AFTER a transformation is the picture of the transformation.
A transformation's preimage is the shape BEFORE the change.

<h3>How do you define relationships between Image and Preimage?</h3>

Usually, the difference between image and pre-image is the way or method of transformation.

<h3>What is transformation in math?</h3>

A transformation is a broad phrase covering four distinct methods of changing the shape and/or location of a point, line, or geometric figure.

The Pre-Image is the original shape of the item, and the Image during the transformation is the final shape and location of the object.

The types of transformation in math are;

translation
rotation
reflection, and
dilation.

Learn more about pre-image:
brainly.com/question/8405245
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7 0

2 years ago