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Lunna [17]
3 years ago
6

Write your answer in simplest form. 11 2/16 - 8 3/16

Mathematics
1 answer:
MariettaO [177]3 years ago
4 0
The answer is 2 15/16.
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Find lim h->0 f(9+h)-f(9)/h if f(x)=x^4 a. 23 b. -2916 c. 2916 d. 2925
Svetach [21]

\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h

Carry out the binomial expansion in the numerator:

(9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4

Then the 9⁴ terms cancel each other, so in the limit we have

\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h

Since <em>h</em> is approaching 0, that means <em>h</em> ≠ 0, so we can cancel the common factor of <em>h</em> in both numerator and denominator:

\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)

Then when <em>h</em> converges to 0, each remaining term containing <em>h</em> goes to 0, leaving you with

\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}

or choice C.

Alternatively, you can recognize the given limit as the derivative of <em>f(x)</em> at <em>x</em> = 9:

f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h

We have <em>f(x)</em> = <em>x</em> ⁴, so <em>f '(x)</em> = 4<em>x</em> ³, and evaluating this at <em>x</em> = 9 gives the same result, 2916.

8 0
3 years ago
HELP ME PLEASE BRAINLIEST ANSWER
Damm [24]

Answer:

D

Step-by-step explanation:

I'm not really 100% sure this is a weird question but I would assume D because if something is not proportional or doesn't conform to an equation, it has no slope

8 0
3 years ago
The number of camera phones shipped globally can be modeled by the function y=1.28e^1.31x where x is the number of years since 1
erik [133]
The first thing you should know for this case is how many years there are from 1997 to 2014.
 The number of years is:
 x = 2014-1997 = 17
 Then, you must replace this value in the given equation:
 y = 1.28e ^ 1.31x
 y = 1.28 * e ^ (1.31 * (17))
 y = 6011018736
 answer:
 the number of camera phones shipped in millions in 2014 is
 y = 6011018736
8 0
3 years ago
The level of nitrogen oxides (NOX) in the exhaust after 50,000 miles or fewer of driving of cars of a particular model varies No
Greeley [361]

Answer:

For a level of 0.0174 or more of nitrogen oxide, the probability of fleet is 0.01.              

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0.02 g/mi

Standard Deviation, σ = 0.01 g/mi

Sample size, n = 81

We are given that the distribution of  level of nitrogen oxides is a bell shaped distribution that is a normal distribution.

Standard error due to sampling:

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.01}{\sqrt{81}} = 0.0011

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.01

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 0.02}{0.0011})=0.01  

= 1 -P( z \leq \displaystyle\frac{x - 0.02}{0.0011})=0.01  

=P( z \leq \displaystyle\frac{x - 0.02}{0.0011})=0.99  

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 0.02}{0.0011} = -2.326\\\\x = 0.0174

For a level of 0.0174 or more of nitrogen oxide, the probability of fleet is 0.01.

5 0
2 years ago
Help please someone I need all answers please
snow_tiger [21]
17. 8/10 is equal to .8 not .08
18a. False
18b. True
18c. False
18d. False
18e. True
19. 0.02
20. No, the zero needs to be in the tenths place or else the decimal (.2) would become 2/10.
21. 17/100
22. Words- Ten hundredths
Fraction- 10/100
Decimal- .10
5 0
3 years ago
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