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3 years ago

Find the Interest Earned.

1 answer:

5 0

A=p(1+i/m)^mn
Interest earned
I=A-p

A=980×(1+0.08÷4)^(4×5)
A=1,456.23

I=1,456.23−980
I=476.23

A=7,200×(1+0.04)^(8)
A=9,853.69

I=9,853.69−7,200
I=2,653.69

A=15,520×(1+0.06÷2)^(2×4)
A=19,660.27

I=19,660.27−15,520
I=4,140.27

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The slope-intercept form of a linear equation is y=mx + b, where x and y are

HACTEHA [7]

Answer:

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Step-by-step explanation:

Given

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6 0

3 years ago

|-15-7|= simply the expression

hichkok12 [17]

Answer:

Step-by-step explanation:

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3 0

3 years ago

Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$

$\boxed{ \tt \red{C}arry \: \red{ O}n \: \red{L}earning} \: \underline{\tt{5/13/22}}$

4 0

2 years ago

The probability that a fair coun tossed 10 times will lsndheads up on all ten tosses is 1 out of ?

boyakko [2]

It is one out of ten

4 0

3 years ago

A scale drawing of a field is shown below. The drawing uses a scale of 1 in. = 14 ft. What is the area of the actual field?

svp [43]

Answer:

D. 4,704 ft²

Step-by-step explanation:

Giving a scale of 1 in. = 14 ft, to find the actual area of the field represented by the triangular drawing above, convert the length of the base and height to the actual lengths of the field using the scale.

Height of drawing = 8 in.

Actual height = 8*14 = 112 ft

Base of drawing = 6 in.

Actual base = 6*14 = 84 ft

Area of the field = area of ∆

Area = ½*base*height

Area = ½*84*112

Area = 4,704 ft²

3 0

2 years ago