Simplify:-3m-1/5(50m+100)-1

First find f′ and then find f. f′′(x)=3x^3+6x^2−x+2, f′(1)=9, f(1)=−7. f′(x)= f(x)=

Vadim26 [7]

I haven't learned antiderivitives yet but I can try to logic it

First find f′ and then find f. f′′(x)=3x^3+6x^2−x+2, f′(1)=9, f(1)=−7.

we reverse chain rule

3x^3, we know that it was a 4th degree thing, and the coefient is 3, so
4*what=3?, answer is 3/4
3/4x^4

6x^2
we know it was x^3, and the coefient is now 6 so
3*what=6? what=2
2x^3

-1x, the power was 2 and coefient is now -1, so
2 times what=-1?, -1/2
-1/2x^2

2, that is from 2x

so

3/4x^4+2x^3-1/2x^2+2x=f'(x)
test x=1
(3/4)(1)+2(1)-(1/2)(1)+2(1)=
3/4+2-2/4+2=
4 and 1/4 we need to get it to 9
4 and 1/4 +what=9
answer is 4 and 3/4
so we add that to the end since it will become 0 from derivitive


f'(x)=3/4x^4+2x^3-1/2x^2+2x+4 and 3/4

now reverse drivitive again

3/4x^4
exponent is 5 and coefient is 3/4
5 times what=3/4? answer is 3/20
3/20x^5

2x^3
exponent should be 4 and coefient is 2
4 times what=2? answer is 1/2
1/2x^4

-1/2x^2
exponent should be 3 and coefient is -1/2
3 times what=-1/2? answer is -1/6
-1/6x^3

2x
exponent should be 2 and coefient is 2
2 times what=2? answer is 1
1x^2

4 and 3/4 turns to (4 and 3/4)x


f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+(4 and 3/4)x
try evaluating it for x=1
f(1)=(3/20)+(10/20)-(10/50)+1+(19/4)
f(1)=6 and 7/30
what do we add to get -7
-13 and 7/30

f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+(4 and 3/4)x-13 and 7/30



ANSWER
f'(x)=3/4x^4+2x^3-1/2x^2+2x+19/4
f(x)=3/20x^5+1/2x^4-1/6x^3+x^2+19/4x-187/30

4 0

3 years ago

large boxes weigh 40 pounds and small boxes weigh 35 pounds, a total of 120 boxes on a truck carrying 4475 pounds altogether . h

klasskru [66]

Do you have answer choices?

7 0

3 years ago

sweet t has 2 orange picks for every five green. if there are 21 picks in all, how many picks are orange

jarptica [38.1K]

21 picks
altogether (2x+5x)=21
7x=21
x=3
2*3=6 orange picks

or
x=orange, y = greens
x+y=21
x/y=2/5,----> y=5x/2
x+5x/2=21
2x/2+5x/2=21
7x/2=21
7x=42,
x=6 orange picks

3 0

3 years ago

Find each of the following products of conjugate pairs.See if you can work out a pattern

yKpoI14uk [10]

Answer: is 0

Step-by-step explanation:

4 0

3 years ago

A cell phone manufacturer claims that the average battery life of its newest flagship smartphone is exactly 20 hours. Javier bel

adelina 88 [10]

Answer:

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours.

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.

Step-by-step explanation:

1) Data given and notation

$\bar X=19.5$ represent the battery life sample mean

$s=1.9$ represent the sample standard deviation

$n=33$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha=0.05$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean battery life is less than 20 :

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

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

$t=\frac{19.5-20}{\frac{1.9}{\sqrt{33}}}=-1.51$

4) P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=33-1=32$

Since is a one-side lower test the p value would be:

$p_v =P(t_{(32)}$

5) Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the average battery life it's not significantly different less than 20 hours at 5% of signficance. If we analyze the options given we have:

(A) Reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. FALSE, we FAIL to reject the null hypothesis.

(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. TRUE, we fail to reject the null hypothesis that the mean would be 20 or higher .

(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours. TRUE, we FAIL to reject the null hypothesis that the mean is greater or equal to 20 hours, so we reject the alternative hypothesis that the mean is less than 20 hours.

(D) There is enough evidence at the α=0.05 level of significance to support the claim that the true population mean battery life of the smartphone is not equal to 20 hours. FALSE the claim is not that the mean is different from 20. The real claim is: "Javier believes the mean battery life is less than 20 hours".

5 0

3 years ago