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4 years ago

Question attached below

Mathematics

1 answer:

tamaranim1 [39]4 years ago

6 0

Answer:

B . $f(x+h)=\frac{x+h}{1+x+h}$

Step-by-step explanation:

Given function is:

$f(x)=\frac{x}{1+x}$

Sall h is used to denote minor change in function. The value of the new function f(x+h) will be calculated by putting x+h in place of x in the function.

So putting x+h in place of x

$f(x+h)= \frac{x+h}{1+(x+h)}\\=\frac{x+h}{1+x+h}$

So, option B is the correct answer ..

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1a) Use the − definition of the limit to prove lim→23+4=10.

True [87]

You're missing symbols in both of your expressions, but considering the first limit has a value of 10, I suspect you meant to write

$\displaystyle \lim_{x\to2}(3x+4) = 10$

(which is true) but unfortunately I am nowhere near as confident about what the second one is supposed to say. So one proof will have to do, unless you come around to editing your question.

The claim,

$\displaystyle \lim_{x\to2}(3x+4) = 10$

is to say that, for any given ε > 0, we can find δ (a number that depends on ε) such that whenever |x - 2| < δ, this ensures that |(3x + 4) - 10| < ε.

Roughly speaking: if x is close enough to 2, this translates to f(x) = 3x + 4 being close enough to 10. It's our job to figure out how close x needs to be to 2 in order that f(x) is close enough to 10, where the closeness to 10 is some given threshold.

We want to arrive at the inequality,

|(3x + 4) - 10| < ε

so suppose we work backwards. With some simplification and rewriting, we have

|3x - 6| = |3 (x - 2)| = |3| |x - 2| = 3 |x - 2| < ε

and so

|x - 2| < ε/3

which suggests that we should pick δ = ε/3.

Now for the proof itself:

Let ε > 0 be given, and let δ = ε/3. Then

|x - 2| < δ = ε/3

3 |x - 2| < ε

|3x - 6| < ε

|(3x + 4) - 10| < ε

and this completes the proof of the limit. QED

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3 years ago

(3cosx-4sinx) + (3sinx+4cosx) = 5

Ksenya-84 [330]

(3 cos x-4 sin x)+(3sin x+4 cos x)=5
(3cos x+4cos x)+(-4sin x+3 sin x)=5
7 cos x-sin x=5
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49 cos²x=25+10 sinx+sin²x
49(1-sin²x)=25+10 sinx+sin²x
49-49sin²x=25+10sinx+sin²x
50 sin² x+10sinx-24=0

Sin x=[-10⁺₋√(100+4800)]/100=(-10⁺₋70)/100
We have two possible solutions:
sinx =(-10-70)/100=-0.8
x=sin⁻¹ (-0.8)=-53.13º (360º-53.13º=306.87)

sinx=(-10+70)/100=0.6
x=sin⁻¹ 0.6=36.87º

The solutions when 0≤x≤360º are: 36.87º and 306.87º.

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3 years ago

when you divide a fraction less than 1 by a whole number greater than 1 is the quotient less than greater than or equal to the d

laiz [17]

The quotient would be less than the dividend because when you divide a fraction less than 1 by a number greater than one, you are dividing a fraction by another term. Any fraction less than 1 divided by a whole number, the equation converted into a multiplication sentence would become the fraction multiplied by the term as the denominator and the numerator as a one, which a fraction multiply a fraction would be a quotient less than the dividend.

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4 years ago

Solve for x: -7 < x-1 < 8 06 0-6 >x>9 06 >x>-9 O-6

avanturin [10]

Answer:

−