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ANEK [815]
3 years ago
8

The rental fee for space at a flea market is $200 per month base rate

Mathematics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

anor soi sena orka uzumo senda?

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A fair spinner has 12 equal sections: 4 red, 3 blue and 5 green.
AnnyKZ [126]
The probability of getting green is-3/12
And not getting green is -12/3
6 0
3 years ago
In a volleyball game, Caroline served the ball over the net 16 out of 20 times. What percent did she not make her serve over the
brilliants [131]

16/20. We reduce that and get 4/5.

After this it's much simpler to chage it.

100 dividied by 5 is 20. (20,40,60,80,100)

So the 4th number (4/100)

is 80. 80%.

20% she didn't and 80% she did :) (100-80)

7 0
3 years ago
Read 2 more answers
Help me asap please!!!
Artemon [7]
Well, just by looking at the beginning of the problem, Jamelia had made the common mistake of thinking that 
\sqrt{72} (<span>8.4852...)
</span>is equal to 
2* \sqrt{36} (12)

If you want to estimate a square root like 72, simply find squares that would fit around the number you are looking to find, in our case, 72.

So 9*9 is 81, which is too high and 8*8 is 64, which is too low. So you know that somewhere between those numbers is what your root of 72 is!
5 0
3 years ago
Write an expression in expanded form and an expression in factored form for the diagram.
olga2289 [7]

Step-by-step explanation:

You put them together to form a bigger number then break it down. sorry if it doesn't make sense.

Btw do you form the number if so I'll help just give the number and I'll break it down for you. :)

6 0
3 years ago
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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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