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loris [4]
3 years ago
12

Can an expert help with this

Mathematics
1 answer:
SVEN [57.7K]3 years ago
7 0

The formula of an area of a circle:

A_O=\pi r^2

We have

r = 1.8 m

π ≈ 3.14

Substitute:

A_O=\pi\cdot1.8^2=3.24\pi\ m^2\approx3.24\cdot3.14=10.1736\ m^2\approx10.17\ m^2

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Multiply and Simplify<br> (x-4)(x^2 - 5x - 6)
kirill [66]

Answer:

x^3 -9x^2 +14x +24

Step-by-step explanation:

(x-4)(x^2 - 5x - 6)

Multiply the x by everything in the second term

x * (x^2 - 5x - 6)

x^3 -5x^2 -6x

Multiply the -4 by everything in the second term

-4 * (x^2 - 5x - 6)

-4x^2 +20x +24

Add everything together

I like to line them up vertically

x^3 -5x^2 -6x

      -4x^2 +20x +24

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

x^3 -9x^2 +14x +24


4 0
2 years ago
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Why am this point be labeled with two names; 1 1/2 and 3/2
kakasveta [241]
3/2 is an improper fraction, where the denominator (bottom number) is smaller than the top (numerator)
1  1/2 is this same fraction but now it is not improper.  
5 0
3 years ago
There are 300 pets in the pet store. 25% of the pets are dogs. How many dogs are in the pet store?
ser-zykov [4K]
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Step-by-step explanation:

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3 years ago
For the function y=x^5+1x^3-30x, find all real zeros.
siniylev [52]

Answer:

The real zeroes are -√5 , 0 , √5

Step-by-step explanation:

* Lets explain how to solve the problem

- The function is y = x^5 + x³ - 30x

- Zeros of any equation is the values of x when y = 0

- To find the zeroes of the function equate y by zero

∴ x^5 + x³ - 30x = 0

- To solve this equation factorize it

∵ x^5 + x³ - 30x = 0

- There is a common factor x in all the terms of the equation

- Take x as a common factor from each term and divide the terms by x

∴ x(x^5/x + x³/x - 30x/x) = 0

∴ x(x^4 + x² - 30) = 0

- Equate x by 0 and (x^4 + x² - 30) by 0

∴ x = 0

∴ (x^4 + x² - 30) = 0

* Now lets factorize (x^4 + x² - 30)

- Let <em>x² = h</em> and <em>x^4 = h²</em> and replace x by h in the equation

∴ (x^4 + x² - 30) = (h² + h - 30)

∵ (x^4 + x² - 30) = 0

∴ (h² + h - 30) = 0

- Factorize the trinomial into two brackets

- In trinomial h² + h - 30, the last term is negative then the brackets

 have different signs (     +     )(     -     )

∵ h² = h × h ⇒ the 1st terms in the two brackets

∵ 30 = 5 × 6 ⇒ the second terms of the brackets

∵ h × 6 = 6h

∵ h × 5 = 5h

∵ 6h - 5h = h ⇒ the middle term in the trinomial, then 6 will be with

  (+ ve) and 5 will be with (- ve)

∴ h² + h - 30 = (h + 6)(h - 5)

- Lets find the values of h

∵ h² + h - 30 = 0

∴ (h + 6)(h - 5) = 0

∵ h + 6 = 0 ⇒ subtract 6 from both sides

∴ h = -6

∵ h - 5 = 0 ⇒ add 5 to both sides

∴ h = 5

* Lets replace h by x

∵ <em>h = x²</em>

∴ x² = -6 and x² = 5

∵ x² = -6 has no value (<em>no square root for negative values</em>)

∵ x² = 5 ⇒ take √ for both sides

∴ x = ± √5

- There are three values of x ⇒ x = 0 , x = √5 , x = -√5

∴ The real zeroes are -√5 , 0 , √5

4 0
3 years ago
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