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2 years ago

calculate the amount of ₹30000 at the end of 2 years 4 months , compounded annually at 10% per annum

Mathematics

1 answer:

marissa [1.9K]2 years ago

6 0

Fv= 30,000 *(1+10%/12)^(2*12+4)
Fv = 30,000*(1+10%/12)^28
Fv= 37,847.46

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Length of the base of the given triangle is (x+3) units and the perpendicular height is (x-1)units. If the area of the triangle

Lana71 [14]

Step-by-step explanation:

the area of a triangle is

baseline × height / 2

so, in our case we know this is 10 :

(x+3)(x-1)/2 = 10

(x+3)(x-1) = 20

x² - x + 3x - 3 = 20

x² + 2x - 23 = 0

that is exactly the provided equation. so, yes, x satisfies it.

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 2

c = -23

x = (-2 ± sqrt(2² - 4×1×-23))/(2×1) =

= (-2 ± sqrt(4 + 92))/2 = (-2 ± sqrt(96))/2 =

= (-2 ± sqrt(16×6))/2 = (-2 ± 4×sqrt(6))/2 =

= -1 ± 2×sqrt(6)

x1 = -1 + 2×sqrt(6) = 3.898979486... ≈ 4

or using sqrt(6) ≈ 2.45

= -1 + 2×2.45 = -1 + 4.9 = 3.9 ≈ 4

x2 = -1 - 2×sqrt(6) = -5.898979486... ≈ -6

or using sqrt(6) ≈ 2.45

= -1 - 2×2.45 = -1 - 4.9 = -5.9 ≈ -6

x2 would give us negative lengths for the triangle, which does not make sense.

so, x = 4 is our solution.

and that makes the height x-1 = 4-1 = 3 units.

7 0

1 year ago

X^2-8x+10=-4 quadratic equation written in standard form

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Rewriting the formula in standard form would be x^2-8x+14=0

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2 years ago

What are the first ten multiples of 12

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12x1
12x2
12x3
12x4
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12x7
12x8
12x9
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4 0

3 years ago

Read 2 more answers

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities