20000=15000(1.072)^t

Answers

Answer 1

Answer:
   20,000 = 15,000 (1.072)^t

Divide each side by 15,000 :    4/3 = (1.072)^t

Take the log of each side:    log(4/3) = t log(1.072)

Divide each side by log(1.072):    t = log(4/3) / log(1.072)

      0.12494 / 0.03019 = 4.138 (rounded)

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Answers

11$ will be your answer

Please someone help me to prove this.

Answers

Answer: see proof below

Step-by-step explanation:

Use the Power Reducing Identity: sin² Ф = (1 - cos 2Ф)/2

Use the Double Angle Identity: sin 2Ф = 2 sin Ф · cos Ф

Use the following Sum to Product Identities:

$\sin x - \sin y = 2\cos \bigg((x+y)/(2)\bigg)\sin \bigg((x-y)/(2)\bigg)\n\n\n\cos x - \cos y = -2\sin \bigg((x+y)/(2)\bigg)\sin \bigg((x-y)/(2)\bigg)$

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \qquad (\sin^2A-\sin^2B)/(\sin A\cos A-\sin B \cos B)$

$\text{Power Reducing:}\qquad (\bigg((1-\cos 2A)/(2)\bigg)-\bigg((1-\cos 2B)/(2)\bigg))/(\sin A \cos A-\sin B\cos B)$

$\text{Half-Angle:}\qquad \qquad (\bigg((1-\cos 2A)/(2)\bigg)-\bigg((1-\cos 2B)/(2)\bigg))/((1)/(2)\bigg(\sin 2A-\sin 2B\bigg))$

$\text{Simplify:}\qquad \qquad (1-\cos 2A-1+\cos 2B)/(\sin 2A-\sin 2B)\n\n\n.\qquad \qquad \qquad =(-\cos 2A+\cos 2B)/(\sin 2A - \sin 2B)\n\n\n.\qquad \qquad \qquad =(\cos 2B-\cos 2A)/(\sin 2A-\sin 2B)$

$\text{Sum to Product:}\qquad \qquad (-2\sin \bigg((2B+2A)/(2)\bigg)\sin \bigg((2B-2A)/(2)\bigg))/(2\cos \bigg((2A+2B)/(2)\bigg)\sin \bigg((2A-2B)/(2)\bigg))$

$\text{Simplify:}\qquad \qquad (-2\sin (A + B)\cdot \sin (-[A - B]))/(2\cos (A + B) \cdot \sin (A - B))$

$\text{Co-function:}\qquad \qquad (2\sin (A + B)\cdot \sin (A - B))/(2\cos (A + B) \cdot \sin (A - B))$

$\text{Simplify:}\qquad \qquad \quad (\cos (A+B))/(\sin (A+B))\n\n\n.\qquad \qquad \qquad \quad =\tan (A+B)$

LHS = RHS: tan (A + B) = tan (A + B) $\checkmark$

Answer:

We know that,

$\dag\bf\:sin^2A=(1-cos2A)/(2)$

$\dag\bf\:sin2A=2sinA\:cosA$

___________________________________

Now, Let's solve !

$\leadsto\:\bf(sin^2A-sin^2B)/(sinA\:cosA-sinB\:cosB)$

$\leadsto\:\sf((1-cos2A)/(2)-(1-cos2B)/(2))/((2sinA\:cosA)/(2)-(2sinB\:cosB)/(2))$

$\leadsto\:\sf(1-cos2A-1+cos2B)/(sin2A-sin2B)$

$\leadsto\:\sf(2sin(2A+2B)/(2)\:sin(2A-2B)/(2))/(2sin(2A-2B)/(2)\:cos(2A+2B)/(2))$

$\leadsto\:\sf(sin(A+B))/(cos(A+B))$

$\leadsto\:\bf{tan(A+B)}$

I put $200 in a saving account with 7% interest a year. How much will i have in 19 years

Answers

it would be 200×0.07= 14×19= 266

I = PRT
P for Principle Amount
R for Rare
T for Time in years

I = 200 × 0.07 × 19 (70% = $(7)/(100)$ = 0.07)
= $266 - this is the interest

You will have 200 + 266 = $466 in your account

How do i find the answer to this problem? 2/6 x 24 = m working with fractions

Answers

Solve for m by simplifying both side of the equation,then isolating the variable.

M=8
Hope that this helps :)

What is the ratio of 33/36

Answers

all you have to do is find a number that both of them is divisible by and divide it by them. for example, when you divide it by 3, you will get 11/12

11/12 Divide both by 3.

5a squared + 4ab
I need help with my online homework

Answers

$5a^2 + 4ab \n \n gcf = a \n \n a( (5a^2)/(a) + (4ab)/(a) ) \n \n a(5a + 4b) \n \n$

The answer is: a(5a + 4b).

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