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Anna [14]
2 years ago
13

One of the factors of the polynomial

Mathematics
1 answer:
oee [108]2 years ago
8 0

Answer:

p = 2/3 while q = 5/3

Hope it helps, cos I'm kinda busy to give the explanation

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How to substitute x=-2 and x=5 into 2x^2+6-20=0 to check my work
Komok [63]
First answer (x=-2)
...
(2)[(-2)^2]+6-20
2(4)+6-20
8+6-20
14-20
<em>-6 (answer)
</em><em>
</em><em />Second answer (x=5)
...
(2)[(-5)^2]+6-20
2(25)+6-20
50+6-20
56-20
<em>36 (answer)
</em>Hope this helps!
4 0
3 years ago
Find the missing side. round to the nearest tenth.
horrorfan [7]

Answer:

24) x = 9.2

25) x = 30.8

Step-by-step explanation:

Given

See attachment for triangles

Solving (24)

To solve for x, we make use of cosine formula

i.e.

cos(40) = adjacent ÷ hypotenuse

So, we have:

cos(40) = x ÷ 12

Multiply both sides by 12

12 cos(40) = x

12 * 0.7660 = x

x = 9.2

Solving (25)

To solve for x, we make use of sine formula

i.e.

sin(25) = opposite ÷ hypotenuse

So, we have:

sin(25) = 13 ÷ x

Multiply both sides by

x sin(25) = 13

Divide by sin(25)

x = 13 ÷ sin(25)

Using a calculator

x = 30.8

5 0
2 years ago
Again ... Commute times in the U.S. are heavily skewed to the right. We select a random sample of 500 people from the 2000 U.S.
VladimirAG [237]

Answer:

We conclude that the mean commute time in the U.S. is less than half an hour.

Step-by-step explanation:

We are given that a random sample of 500 people from the 2000 U.S. Census is selected who reported a non-zero commute time.

In this sample the mean commute time is 27.6 minutes with a standard deviation of 19.6 minutes.

Let \mu = <u><em>mean commute time in the U.S..</em></u>

So, Null Hypothesis, H_0 : \mu \geq 30 minutes      {means that the mean commute time in the U.S. is more than or equal to half an hour}

Alternate Hypothesis, H_A : \mu < 30 minutes     {means that the mean commute time in the U.S. is less than half an hour}

The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;

                           T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean commute time = 27.6 minutes

            s = sample standard deviation = 19.6 minutes

            n = sample of people from the 2000 U.S. Census = 500

So, <u><em>the test statistics</em></u>  =  \frac{27.6 -30}{\frac{19.6}{\sqrt{500} } }  ~ t_4_9_9

                                       =  -2.738

The value of t test statistic is -2.738.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the t table gives critical values of -1.645 at 499 degree of freedom for left-tailed test.</u>

Since our test statistic is less than the critical value of t as -2.378 < -1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis.</u>

Therefore, we conclude that the mean commute time in the U.S. is less than half an hour.

4 0
3 years ago
Change the following subtraction expression to addition: 2 - 11
tester [92]

Answer:

I think it would be 2+ -11

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f
tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :-  f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x)  to have a maximum or minimum at given point.

ii)  find first derivative f^{l} (x) and equating zero

iii) solve and find 'x' values

iv) Find second derivative f^{ll}(x) >0 then find the minimum value at x=a

v) Find second derivative f^{ll}(x) then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

<u>step1:</u>- find first derivative f^{l} (x) and equating zero

  f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)

f^{l}(x) = \frac{1}{x^2+1} (2x)  ……………(1)

f^{l}(x) = \frac{1}{x^2+1} (2x)=0

the point is x=0

<u>step2:-</u>

Again differentiating with respective to 'x', we get

f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}

on simplification , we get

f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}

put x= 0 we get f^{ll}(0) = \frac{2}{(1)^2}   > 0

f^{ll}(x) >0 then find the minimum value at x=0

<u>Final answer</u>:-

The minimum value of the given function is f(0) = 0

5 0
3 years ago
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