Given that AB = 370 m and AC = 510 m, find
(i) the distance between B and C,
(ii) ACB,
(iii) the bearing of C from B,
(iv) the shortest distance from A to BC.​

Given That AB = 370 M And AC = 510 M, Find(i) The Distance Between B And C,(ii) ACB,(iii) The Bearing

Answers

Answer 1

Answer:

(i) The distance between B and C is approximately 552.9 m

(ii) ∠ACB is approximately 40.49°

(iii) The bearing of C from B is approximately 184.49°

(iv) The shortest distance from A to BC is approximately 331.155 m

Step-by-step explanation:

The given parameters are;

The length of segment AB = 370 m

The length of segment AC = 510 m

The bearing of B from A = 68°

The bearing of C from A = 144°

(i) From the bearing of B from A = 68° and the bearing of C from A = 144°, we have;

∠BAC = 144° - 68° = 76°

∠BAC = 76°

Let A represent ∠BAC, c represent segment AB, b represent segment AC, and a represent segment BC, by cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

By substituting the known values, we get;

a² = 510² + 370² - 2 × 510 × 370 × cos(76°)

a² ≈ 397000 - 91301.3234 ≈ 305698.677

BC = a ≈ √(305,698.677) ≈ 552.9

The distance between B and C, BC ≈ 552.9 m

(ii) By sine rule, we have;

a/(sin(A) = b/(sin(B)) - c/(sin(C))

Therefore;

552.9/(sin (76°)) = 370/(sin(C))

sin(C) = 370/(552.9/(sin (76°))) ≈ 0.649321

C = arcsine(0.649321) ≈ 40.49°

∠C = ∠ACB ≈ 40.49°

(iii) Angle ∠B = ∠ABC = 180° - 76° - 40.49° ≈ 63.51°

The bearing of A from B = 360° - (180° - 68°) = 248°

Therefore, the bearing of C from B ≈ 248° - 63.51° ≈ 184.49°

(iv) The shortest distance from A to BC = 370 m × sin(63.51°) ≈ 331.155 m


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Answers

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Nearly all of those answers involve a fraction form and an equivalent decimal form.

===========================================

Work Shown:

Part A

The largest number in the bag is 50, so we cannot select anything larger than that. The probability of getting something larger than 50 is 0% which converts to the decimal form 0

----------------------------

Part B

List out the multiples of 5 to get {5,10,15,20,25,30,35,40,45,50}

Note how 50/5 = 10, which shows there are 10 multiples listed above.

We have 10 items in that set out of 50 items in the set {1,2,3,...,49,50}. The probability of getting a multiple of 5 is 10/50 = 1/5. This converts to the decimal form 0.2

----------------------------

Part C

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There are 15 items in that list out of 50 items in the bag, so 15/50 = (5*3)/(5*10) = 3/10 is the answer. Converting to decimal form gets us 3/10 = 0.3

----------------------------

Part D

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Answers

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Answers

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Step-by-step explanation:

Step-by-step explanation:

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With full process ​

Answers

Answer:

[tex]\displaystyle \frac{d}{dx} \Big [ \frac{3}{x^2} \Big ]=-\frac{6}{x^3}[/tex]

Step-by-step explanation:

We can use either the Power Rule or the Quotient Rule to find the derivative of 3/x².

Power Rule: [tex]\displaystyle \frac{d}{dx}[ x^n] = nx^n^-^1[/tex]  

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Quotient Rule: [tex]\displaystyle \frac{d}{dx} \Big [ \frac{f(x)}{g(x)} \Big ]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}[/tex]

Substituting 3 for f(x) and x² for g(x), we get:

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[tex]\displaystyle \frac{-6x}{x^4}[/tex]

Using exponent rules, we can rewrite this as:

[tex]-6x^1^-^4 = -6x^-^3[/tex]

This can be rewritten as [tex]\displaystyle -\frac{6}{x^3}[/tex], which is the same as what we got using the product rule.

Answer:

[tex]\displaystyle \frac{d}{d x}\left[\frac{3}{x^2}\right] = -\frac{6}{x^{3}}[/tex].

Step-by-step explanation:

Let [tex]f(x)[/tex] and [tex]g(x)[/tex] denote two functions of [tex]x[/tex]. Assume that [tex]g(x) \ne 0[/tex]. The quotient rule states that:

[tex]\displaystyle \frac{d}{d x}\left[\frac{f(x)}{g(x)}\right] = \frac{{f}^\prime(x) \cdot g(x) - f(x) \cdot g^{\prime}(x)}{{(g(x))}^2}[/tex].

In this question:

The numerator of the fraction is [tex]f(x) = 3[/tex] (a constant function.)The denominator of the fraction is [tex]g(x) = x^{2}[/tex].

Find [tex]{f}^{\prime}(x)[/tex] and [tex]{g}^{\prime}(x)[/tex].

Notice that the value of [tex]f(x)[/tex] is constantly [tex]3[/tex] regardless of the value of [tex]x[/tex]. By the constant rule, [tex]{f}^{\prime}(x) = 0[/tex].

For [tex]{g}^{\prime}(x)[/tex], consider the power rule:

if [tex]m[/tex] represents a rational number (such as [tex]2[/tex],) then by the power rule, [tex]\displaystyle \frac{d}{d x}\left[{x}^{m}\right] = m\, {x}^{m-1}[/tex].

Apply this rule to find [tex]{g}^{\prime}(x)[/tex].

[tex]\begin{aligned}{g}^{\prime}(x) &= \frac{d}{d x} \left[x^{2}\right] \\ &= 2\, x^{2 - 1} \\ &= 2\, x\end{aligned}[/tex].

Substitute [tex]f(x) = 3[/tex], [tex]{f}^{\prime}(x) = 0[/tex], [tex]g(x) = x^{2}[/tex], and [tex]{g}^{\prime}(x) = 2\, x[/tex] into the quotient rule expression to find [tex]\displaystyle \frac{d}{d x}\left[\frac{3}{{x}^{2}}\right][/tex]:

[tex]\begin{aligned}&\; \frac{d}{d x}\left[\frac{3}{{x}^{2}}\right] \quad \genfrac{}{}{0em}{}{\leftarrow f(x) = 3}{\leftarrow g(x) = {x}^{2}} \\ =&\; \frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]\\ =&\; \frac{{f}^\prime(x) \cdot g(x) - f(x) \cdot g^{\prime}(x)}{{(g(x))}^2} \\ =& \; \frac{0 \, x^2 - 3\, (2\, x)}{\left({x}^{2}\right)} \\ =&\; -\frac{6}{x^{3}}\end{aligned}[/tex].

Therefore, [tex]\displaystyle \frac{d}{d x}\left[\frac{3}{{x}^{2}}\right] = -\frac{6}{x^{3}}[/tex].

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Answers

Answer:

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Step-by-step explanation:

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