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Rama09 [41]
3 years ago
6

You have a square garden with side lengths of 6 meters. How many square meters is the garden

Mathematics
1 answer:
kompoz [17]3 years ago
4 0

Answer:

Im 100% sure its 24

Step-by-step explanation:

the answer is 24 because if you have a square all the sides are equal and it says that the sides are 6 square meters and a square has 4 sides so 6 times 4 is 24

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What is the factor for this expression:<br><br> 4x+20
sveticcg [70]

Answer:

4(x+5)

Step-by-step explanation:

factor out the 4

4(x+5)

7 0
3 years ago
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How many feet are in 84 yards. (im really bad at converting) :(
Gre4nikov [31]
1 yard = 3 feet, so we would do 3×84, which is 252.

There are 252 feet in 84 yards.
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3 years ago
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Which graph shows a function whose inverse is also a function?
strojnjashka [21]

Answer:

A

Step-by-step explanation:

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3 years ago
Which expressions are equivalent to √/32?<br> 4√8<br> 08<br> 2√8<br> 4√2<br> 16<br> 16√2
vaieri [72.5K]

Answer: 4√2

square root of a number is the factor that we can multiply by itself to get that number.

The square root symbol is also called as a RADICAL.

Perfect squares are the squares of the integers also called as square numbers.

to find the square root of non-perfect squared numbers we have to check the number with a near-perfect squared numbers

how to find square root of non perfect squared numbers:

If we can't figure out what factor multiplied by itself will result in the given number, we can make a factor tree.

Factor tree of 32 = 2*2*2*2*2

                            = (2*2) * (2*2) *2

                            = (4)2 *2

Therefore √/32 = √/2*2*2*2*2

                          = √/(4)2*2

                          = 4√/2

learn more about square roots here:

brainly.com/question/428672

brainly.com/question/3617398

#SPJ9

8 0
2 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
3 years ago
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