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Inessa05 [86]
3 years ago
8

What is the result of dividing 2x3 – 3x2 – 17x + 30 by x + 3

Mathematics
1 answer:
jekas [21]3 years ago
7 0
2x^3 - 3x² - 17x + 30 | x + 3 
<span>________________________ </span>
2x^3 + 6x² _________| 2x² - 9x + 10 
<span>_____-9x² - 17x ____ | </span>
<span>_____-9x² - 27x ____ | </span>
<span>_________ 10x + 30 _| </span>
<span>_________ 10x + 30 _| </span>
<span>________ _____ __0 _| 
</span>
Ans = 2x² - 9x + 10

BRAINLIEST PLS!!!
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I need help , fast quick
Agata [3.3K]

Answer:

33

Step-by-step explanation:

So if u divide each number of players by the number of winners each time u come out with 3.something

So if ur looking for 100 players ud want the number of winners divided by that to equal 3.something

So 100/33=3.something

Plus every answer goes up so u also know its true

Hope this helped

Plz mark Brainliest ;p

8 0
3 years ago
For the pair of functions, find the indicated sum, difference, product, or quotient. f(x) = 16 - x2; g(x) = 4 - x Find (f + g)(x
Rudik [331]
(16-x²)+(4-x) or (-x²+16)+(-x+4)

Combine like terms

(-x²+20-x) or (-x²-x+20)
7 0
3 years ago
Read 2 more answers
Men are believed to have a slightly longer average length of short hospital stays than women; 5.2 days versus 4.5 days respectiv
Georgia [21]

Answer:

Since our calculated value is lower than our critical value,z_{calc}=2.02, we have enough evidence to FAIL to reject the null hypothesis at 1% of singificance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

Step-by-step explanation:

1) Data given and notation  

\bar X_{1}=5.2 represent the mean for group men  

\bar X_{2}=4.5 represent the mean for group women  

Assuming these values for the remaining data:

\sigma_{1}=1.2 represent the population standard deviation for the sample men

\sigma_{2}=1.5 represent the population standard deviation for the sample women

n_{1}=32 sample size for the group men  

n_{2}=30 sample size for the group women  

z would represent the statistic (variable of interest)

p_v represent the p value  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:  

H0:\mu_{1} \leq \mu_{2}  

H1:\mu_{1} > \mu_{2}  

If we analyze the size for the samples both are higher than 30, and we know the population deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:  

z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

3) Calculate the statistic  

We have all in order to replace in formula (1) like this:  

z=\frac{5.2-4.5}{\sqrt{\frac{1.2^2}{32}+\frac{1.5^2}{30}}}=2.02  

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a right tailed test, so we are looking on the normal standard distribution a value that accumulats 0.01 of the area on the right and 0.99 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.01,0,1)", and we got z_{critical}=2.33

5) Statistical decision

Since our calculated value is lower than our critical value,z_{calc}=2.02, we have enough evidence to FAIL to reject the null hypothesis at 1% of significance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

6 0
3 years ago
What are m∠1 and m∠2?
tankabanditka [31]

Answer:

Not enough information

4 0
2 years ago
Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A
Vlad [161]

Answer:

y(x)=sin(2x)+2cos(2x)

Step-by-step explanation:

y''+4y=0

This is a homogeneous linear equation. So, assume a solution will be proportional to:

e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda

Now, substitute y(x)=e^{\lambda x} into the differential equation:

\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0

Using the characteristic equation:

\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0

Factor out e^{\lambda x}

e^{\lambda x}(\lambda ^2 +4) =0

Where:

e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda

Therefore the zeros must come from the polynomial:

\lambda^2+4 =0

Solving for \lambda:

\lambda =\pm2i

These roots give the next solutions:

y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}

Where c_1 and c_2 are arbitrary constants. Now, the general solution is the sum of the previous solutions:

y(x)=c_1 e^{2ix} +c_2 e^{-2ix}

Using Euler's identity:

e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)

y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\

Redefine:

i(c_1-c_2)=c_1\\\\c_1+c_2=c_2

Since these are arbitrary constants

y(x)=c_1sin(2x)+c_2cos(2x)

Now, let's find its derivative in order to find c_1 and c_2

y'(x)=2c_1 cos(2x)-2c_2sin(2x)

Evaluating    y(0)=2 :

y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2

Evaluating     y'(0)=2 :

y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1

Finally, the solution is given by:

y(x)=sin(2x)+2cos(2x)

5 0
3 years ago
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