The amount of money in an account with continuously compounded interest is given by the formula A=Pe^rt, where P is the principal, r is the annual interest rate, and t is the time in years. Calculate to the nearest hunderdth of a year how long it takes for an amount of the money to double if the interest is compounded continuously at 6.2 %. Round to the nearest tenth

Answers

Answer 1

Answer: 2p=pe^0.062t
2=e^0.062t
0.062t = ln2
t = 11.2 years

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A company makes a product and has no way to determine which ones are faulty until an unhappy customer returns it. Three percent of the products are faulty and will cost the company $200 each in customer service and repairs. If the company does not refund the customer when repairing the item, how much should the company charge to make a profit of $2.00 per item? $6.00 $6.19 $8.00 $8.25

Answers

(x) - (0.03)(200) = 2.00
Let x be the company charge to make the profit

solving for x gives 8$

Answer: $8.00

Step-by-step explanation:

Given: Percent of products which are faulty=3%=0.03

Cost of each faulty product =$200

Let x be the company charge to make the profit .

According to the question,

$x-0.03*200=2.00\n\Rightarrow\ x-6.00=2.00\n\Rightarrow\ x=2.00+6.00\n\Rightarrow\ x=$8.00$

Hence, the company should charge $ 8.00 to make a profit of $2.00 per item.

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 !

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

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

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

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

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

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

If you divided 48 objects into sets of 7, how many sets of 7 could you make, and how many are left over?

Answers

Answer:

Step-by-step explanation:

No. objects in one set = 7

Let there be n set

Then, No. objects in n set = 7*n = 7n

Given that total no of objects are 48

Thus,

7n = 48

n = 48/7 = (42+6)/7 = 42/7 + 6/7 = 6 + 6/7

n = 6 + 6/7

Thus, there will be 6 set which has 7 objects. fraction means one set is fraction 6/7 but set cannot be in fraction hence negating it.

Now, total objects in 6 sets = 6*7 = 42.

No. of objects left = 48 - 6 = 6

What's (5xty)-(4x-9)??? And what's 33.7y+8.4-2.04y??? And 4.2x+8.1x+1.8x-2.1x??? And x+2(x-y)???

Answers

You're first problem is unclear as I am not sure if that t was suppose to be there or not.

but as for the second one (33.7y+8.4-2.04y) we would first have to combine like terms so you'd subtract 2.04y from 33.7y , which would give you ; 33.66y so now your answer would be " 33.66y + 8.4 ",
there's nothing more you can do because they're seperate terms c: .

Your third problem ( 4.2x+8.1x+1.8x-2.1x ) is also combining like terms. Add the first 3 terms and that'll give you 14.1x , now subtract 2.1x and that'll give you "12x" as your final answer ~

For your fourth problem x+2 ( x-y ) , you would have to do something called factoring , basically all you gotta do is distribute term by term. So, x times x would give you x^2 , x times -y would give you -xy , 2 times x is 2x , and 2 times -y is -2y , so that'll look somethng like x^2 - xy + 2x - 2x . Now the 2x's cancel out
and you are left with "x^2 - xy" as your final answer ~

A subtending arc on a circle with a radius of 4.5 centimeters has an arc length of 8π. The measure of the angle subtended by the arc is

Answers

The arc length is 8π.
The circumference of the circle is 2πr = 9π.

$(arc)/(circumference) = (angle)/(360)$

so $(8 \pi )/(9 \pi ) = (x)/(360)$

cross-multiply: 9x = 360(8)

x = 320

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