Add. 2.9 x 105 8.7 x 105

If I invest 5,000 at a rate 8.5% for 13 days? How much interest will I earn on my investment if the interest is compounded daily

bogdanovich [222]

Using the formula A = P(1 + r/n)^nt

P = 5,000

r = 8.5% or 0.085

n = 365

t = 13/365 (13 days out of a year)

A = 5,000 (1 + 0.085/365)^365*13/365

A = 5015.16

You will earn $15.16

4 0

3 years ago

Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)

Allisa [31]

Answer:

Given that we have;

$sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }$

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) } =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

$\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

From trigonometric identities, we have;

$sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }$

$\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}$

$1 - sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}$

$1 + sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}$

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) } =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

$\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}$

3 0

3 years ago

I need help with my IXL long terms. Can someone help/tell me how to do them????

ivanzaharov [21]

The answer to this is y=x+5

8 0

3 years ago

A cone-shaped paper water cup has a height of 12 cm and a radius of 6 cm. If the cup is filled with water to one-sixth its hei

Liono4ka [1.6K]

Answer:

1/216

Step-by-step explanation:

The ratio of volumes of similar shapes is the cube of the scale factor. The filled portion of the cone is similar to the entire cone.

<h3>Linear scale factor</h3>

If the filled height is 1/6 of the total height, the scale factor for linear dimensions is 1/6.

<h3>Volume scale factor</h3>

The scale factor for the volume is the cube of the scale factor for linear dimensions. it is ...

(1/6)³ = 1/216

The volume of the filled portion of the cup is 1/216 of the volume of the cup.

8 0

2 years ago

Simplify the expression by combining like terms. 15+12−5+4−7

Fiesta28 [93]

Answer:

31-12

Step-by-step explanation:

15+12-5+4-7

27-5+4-7

31-5-7

31-12

7 0

3 years ago

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