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anyanavicka [17]
2 years ago
7

Help. I need help with these questions ( see image). Please show workings.

Mathematics
1 answer:
Andrew [12]2 years ago
3 0

9514 1404 393

Answer:

  4)  6x

  5)  2x +3

Step-by-step explanation:

We can work both these problems at once by finding an applicable rule.

  \text{For $f(x)=ax^n$}\\\\\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}\dfrac{a(x+h)^n-ax^n}{h}\\\\=\lim\limits_{h\to 0}\dfrac{ax^n+anx^{n-1}h+O(h^2)-ax^n}{h}=\boxed{anx^{n-1}}

where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.

This can be referred to as the <em>power rule</em>.

Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:

  lim[h→0](f(x+h)-f(x))/h = 2ax +b

__

4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.

5. The gradient of x^2 +3x +1 is 2x +3.

__

If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.

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if 7 &lt; 5, then 5 &lt; 3 state the inverse, contrapositive, and converse of each conditional statement. then indicate whether
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The given conditional statement is : If 7<5, then 5<3.

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From the properties of conditional statement we know that:

if p→q then

  • inverse is : \backsim p \rightarrow \backsim q
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Let us elaborate on the above conditions. Let us take two statement p and q  such that If p happens then q also happens.

  • The inverse of a conditional statement is given as "if not p then not q"the converse is given as "if q then p"The contrapositive of the statement states "if not q then not p"
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To learn more about conditional statements:

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Urgent, It is a Calculus question and I’ll appreciate your help. Thanks
BaLLatris [955]

Answer:

  4733

Step-by-step explanation:

Please refer to the attached diagram.

Point A can be assigned x-coordinate "p". Then its y-coordinate is 6p^2. The slope at that point is y'(p) = 12p.

Point B can be assigned x-coordinate "r". Then its y-coordinate is 6r^2. The slope at that point is y'(r) = 12r.

We want the slopes at those points to have a product of -1 (so the tangents are perpendicular). This means ...

  (12p)(12r) = -1

  r = -1/(144p)

The slope of line AB in the diagram is the ratio of the differences of y- and x-coordinates:

  slope AB = (ry -py)/(rx -px) = (6r^2 -6p^2)/(r -p) = 6(r+p) . . . . simplified

The slope of AB is also the tangent of the sum of these angles: the angle AC makes with the x-axis and angle CAB. The tangent of a sum of angles is given by ...

  tan(α+β) = (tan(α) +tan(β))/1 -tan(α)·tan(β))

__

Of course the slope of a line is equal to the tangent of the angle it makes with the x-axis. The tangent of angle CAB is 2 (because the aspect ratio of the rectangle is 2). This means we can write ...

  slope AB = ((slope AC) +2)/(1 -(slope AC)(2))

  6(p+r)=\dfrac{12p+2}{1-(12p)(2)}\\\\3(p+r)(1-24p)=6p+1\qquad\text{multiply by $1-24p$}\\\\3\left(p-\dfrac{1}{144p}\right)(1-24p)=6p+1\qquad\text{use the value for r}\\\\3(144p^2-1)(1-24p)=144p(6p+1)\qquad\text{multiply by 144p}\\\\ 3456 p^3+ 144 p^2+ 24 p+1 =0\qquad\text{put in standard form}\\\\144p^2(24p+1)+(24p+1)=0\qquad\text{factor by pairs}\\\\(144p^2+1)(24p+1)=0\qquad\text{finish factoring}\\\\p=-\dfrac{1}{24}\qquad\text{only real solution}\\\\r=\dfrac{-1}{144p}=\dfrac{1}{6}

So, now we can figure the coordinates of points A and B, and the distance between them. That distance is given by the Pythagorean theorem as ...

  d^2 = (6r^2 -6p^2)^2 +(r -p)^2

  d^2 = (6(1/6)^2 -6(-1/24)^2)^2 +(1/6 +1/24)^2 = 25/1024 +25/576 = 625/9216

Because of the aspect ratio of the rectangle, the area is 2/5 of this value, so we have ...

  Rectangle Area = (2/5)(625/9216) = 125/4608 = a/b

Then a+b = 125 +4608 = 4733.

_____

<em>Comment on the solution</em>

The point of intersection of the tangent lines is a fairly messy expression, and that propagates through any distance formulas used to find rectangle side lengths. This seemed much cleaner, though maybe not so obvious at first.

6 0
3 years ago
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