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postnew [5]
3 years ago
5

HELP ME PLEASEEEEEEEEEEEEEE

Mathematics
2 answers:
CaHeK987 [17]3 years ago
7 0
I think x=24 but not sure
barxatty [35]3 years ago
6 0

Answer:

x = 19 ft

Step-by-step explanation:

\frac{x}{57}  =  \frac{8}{24} \\  \\  \frac{x}{57}  =  \frac{1}{3}  \\  \\ x =  \frac{57}{3}  \\  \\ x = 19\: ft

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Halp mi now. pls<br> pls plsp lls plsp pls
Thepotemich [5.8K]

Answer:

-r+5<-3

Step-by-step explanation:

plug in 19 to the equation inequality

7 0
2 years ago
At the beginning of the year there were 25 students in our
Alborosie
The answer is a 12% increase. 
8 0
3 years ago
How are fractions and decimals related?
Savatey [412]
Fractions and decimals are related in a very close way. Ways such as they are both representing a whole number. 

for example 

1/2 , 0.5 , 50% 

all three numbers have the same value but are expressed in a different format. all three represent the same whole number. 

another example would be

1/4 , 0.25 , 25% 

hopefully this helped. and or makes sense. 

LET ME KNOW :p



7 0
3 years ago
Can someone explain this? and what are the answers- :'_
Sunny_sXe [5.5K]

Answer:

1) b=-1.6 2) g=3 3) x=0.8 4) w=4.9 5) p=-16.2 6) y=9.2

Step-by-step explanation:

Use inverese operations to isolate variables ex. x+4=2

x times 2 = 6                                                            -4

/2                                                                               x=-2

x=3                                                                        

                                                                               

4 0
3 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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