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aniked [119]
3 years ago
8

How do u factor this

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
7 0

You can factor a parabola by finding its roots: if

p(x)=x^2+bx+c

has roots x_1,\ x_2, then you have the following factorization:

p(x) = (x-x_1)(x-x_2)

In order to find the roots, you can use the usual formula

x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

In the first example, this formula leads to

x_{1,2} = \dfrac{-2\pm\sqrt{4+4}}{2} = \dfrac{-2\pm\sqrt{8}}{2} = \dfrac{-2\pm2\sqrt{2}}{2} = 1 \pm \sqrt{2}

So, you can factor

x^2-2x-1 = (x-1-\sqrt{2})(x-1+\sqrt{2})

The same goes for the second parabola.

As for the third exercise, simply plug the values asking

f(1.5)=-5.25

you get

f(-1.5) = 1.5c-3 = -5.25

Add 3 to both sides:

1.5c = -2.25

Divide both sides by 1.5:

c = 1.5

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The sum of n terms of a geometric series is given by
S_{n}=a_{1}\dfrac{r^{n}-1}{r-1}

Substituting the given numbers, you have
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4 0
3 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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3 years ago
The amount spent on heating depends on the average temperature
anastassius [24]

That's D.


A(t) is the amount as a function of the temperature t.



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At the beginning of the week, Aksa’s parents deposited $20 into Aksa’s lunch account. The amount of money Aksa had left after ea
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A number to the 10th power divided by the same number to the 7th power equals 125. What is the number?
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Answer:

24

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7 times 7 is 24

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