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3 years ago

PLEASE HELPP PLEASE find the zeros and factored form of c(x)=3x^2-3

Mathematics

2 answers:

a_sh-v [17]3 years ago

6 0

Answer:

see explanation

Step-by-step explanation:

Given

c(x) = 3x² - 3 ← factor out 3 from each term

= 3(x² - 1) ← x² - 1 is a difference of squares

= 3(x - 1)(x + 1) ← in factored form

To find the zeros equate c(x) to zero, that is

3(x - 1)(x + 1) = 0

Equate each factor to zero and solve for x

x + 1 = 0 ⇒ x = - 1

x - 1 = 0 ⇒ x = 1

zeros are x = - 1, x = 1

Zepler [3.9K]3 years ago

4 0

Answer:

3 ( x-1) (x+1) is the factored form

zeros are x=1, -1

Step-by-step explanation:

3x^2-3

Factor out a 3

3 ( x^2 -1)

What is inside the parentheses is the difference of squares

( a^2 -b^2) = ( a-b)(a+b)

3 ( x-1) (x+1)

To find the zeros, set this equal to zero

3 ( x-1) (x+1) =0

Using the zero product property

x-1 =0 x+1 =0

x=1, x=-1

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Answer:

y = 6·sin(3·(x - 1)) + c

Step-by-step explanation:

The general form of an equation for a sinusoidal function is presented ad follows;

y = a·sin(b·(x - h) + c

Where;

a = The amplitude of the equation

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h = The phase shift

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From the question, we have;

a = 6,

2·π/3 = 2·π/b

∴ b = 3

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We get;

y = 6·sin(3·(x - 1)) + c.

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3 years ago

A collection of 108 coins containing only quarters and nickels is worth $21.

Rudiy27

Answer:

The value could replace q is 108 - n ⇒ the last answer

Step-by-step explanation:

* Lets study the values of quarters and nickles

- One quarter = 25 cents

- One nickle = 5 cents

- The number of coins is 108

- The coins are nickles or quarters only

- The coins worth $ 21

* Lets solve the problem

∵ The number of coins is 108 coins

- Let the quarter is q and the nickel is n

∴ q + n = 108 ⇒ (1)

∵ The coins worth $21

∵ $ 1 = 100 cents

∴ $21 = 21 × 100 = 2100 cents

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∵ The nickels = 5 cents

∴ 25q + 5n = 2100 ⇒ (2)

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∵ q + n = 108 ⇒ subtract n from both sides

∴ q = 108 - n

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3 years ago

Read 2 more answers

A spherical balloon was inflated such that it had a diameter of 24 centimeters. Additional helium was then added to the balloon

DanielleElmas [232]

To solve this we are going to use the formula for the volume of a sphere: $V= \frac{4}{3} \pi r^3$
where
$r$ is the radius of the sphere

Remember that the radius of a sphere is half its diameter; since the first radius of our sphere is 24 cm, $r= \frac{24}{2} =12$ . Lets replace that in our formula:
$V= \frac{4}{3} \pi r^3$
$V= \frac{4}{3} \pi (12)^3$
$V=7238.23 cm^3$

Now, the second diameter of our sphere is 36, so its radius will be: $r= \frac{36}{2} =18$ . Lets replace that value in our formula one more time:
$V= \frac{4}{3} \pi r^3$
$V= \frac{4}{3} \pi (18)^3$
$V=24429.02$

To find the volume of the additional helium, we are going to subtract the volumes:
Volume of helium= $24429.02cm^3-7238.23cm^3=17190.79cm^3$

We can conclude that the volume of additional helium in the balloon is approximately <span>17,194 cm³.</span>

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