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3 years ago

I need to know how to do simple interest

2 answers:

8 0

Use this simple interest calculator to find A, the Final Investment Value, using the simple interest formula: A = P(1 + rt) where P is the Principal amount of money to be invested at an Interest Rate R% per period for t Number of Time Periods. Where r is in decimal form; r=R/100; r and t are in the same units of time.

REY [17]3 years ago

5 0

Answer:

Step-by-step explanation: Simple Interest Equation (Principal + Interest)

A = P(1 + rt)

Where:

A = Total Accrued Amount (principal + interest)

P = Principal Amount

I = Interest Amount

r = Rate of Interest per year in decimal; r = R/100

R = Rate of Interest per year as a percent; R = r * 100

t = Time Period involved in months or years

From the base formula, A = P(1 + rt) derived from A = P + I and I = Prt so A = P + I = P + Prt = P(1 + rt)

Note that rate r and time t should be in the same time units such as months or years. Time conversions that are based on day count of 365 days/year have 30.4167 days/month and 91.2501 days/quarter. 360 days/year have 30 days/month and 90 days/quarter.

Simple Interest Formulas and Calculations:

The accrued amount of an investment is the original principal P plus the accumulated simple interest, I = Prt, therefore we have:

A = P + I = P + (Prt), and finally A = P(1 + rt)

Calculate Total Amount Accrued (Principal + Interest), solve for A

A = P(1 + rt)

Calculate Principal Amount, solve for P

P = A / (1 + rt)

Calculate rate of interest in decimal, solve for r

r = (1/t)(A/P - 1)

Calculate rate of interest in percent

R = r * 100

Calculate time, solve for t

t = (1/r)(A/P - 1)

This is better...

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Please help me prove this theorem thing. I'll give yuh mind kisses <3 have a good day luvlies

lesantik [10]

Answer:

∠ 1 = 20 x + 5

∠ 2 = 24 x - 1

If lines l and m are parallel:

∠ 1 + ∠ 2 = 180°

20 x + 5 + 24 x - 1 = 180

44 x + 4 = 180

44 x = 180 - 4

44 x = 176

x = 176 : 44

x = 4°

Your welcome again :)

Step-by-step explanation:

6 0

3 years ago

Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: Find the volume change

$\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$