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2 years ago

If a person walks 1/3 of a mile every 1/6 of an hour, what is their speed in miles per hour?

Mathematics

2 answers:

zhannawk [14.2K]2 years ago

4 0

Answer:

2 mi per hour

Step-by-step explanation:

1/3*6=

6/3= 2

2 mi per hour

Dima020 [189]2 years ago

4 0

Answer:

2mph

Step-by-step explanation:

If a person walks 1/3 of a mile every 1/6 of an hour you divide to solve.

$\frac{\frac{1}{3}}{\frac{1}{6}}$ To solve you flip the bottom and multiply:

$\frac{1}{3} *\frac{6}{1}$ = $\frac{6}{3} = 2$

The person is walking 2mph

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siniylev [52]

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Step-by-step explanation:

(5+x)+12

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3 years ago

a cookie jar contains four oatmeal raisin, one sugar, nine chocolate chip, and six peanut butter cookies. if one is chosen at ra

VikaD [51]

Step-by-step explanation:

There are a total of 4 + 1 + 9 + 6 = 20 cookies. So the probabilities of each type for a random cookie are:

P(oatmeal raisin) = 4/20 = 1/5

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3 years ago

Captain's Autos sells 22 used cars on

aliina [53]

Answer:

160

Step-by-step explanation:

22 + 18 = 25 percent

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1 year ago

Mr. Charles gave both of his sons, Joe and Bill, $5,000. Joe deposited the money in an account that offered simple interest of 1

Maksim231197 [3]

Answer:

Bill by $85

Step-by-step explanation:

Bill : Compound interest is calculated by multiplying the initial principal amount by one plus the annual interest rate raised to the number of compound periods minus one. The total initial amount of the money given is then subtracted from the resulting value.

Compounded annually for 6 years at 9 % interest Bill would have $8385.50.

Joe : Simple interest at 11% a year The formula we'll use for this is the simple interest formula.

P is the principal amount, $5000.00.

r is the interest rate, 11% per year, or in decimal form, 11/100=0.11.

t is the time involved, 6year(s) time periods.

So, t is 6year time periods.

To find the simple interest, we multiply 5000 × 11 × 6 = 3300

Usually now, the interest is added onto the principal to figure some new amount after 6 year(s),

or 5000.00 + 3300.00 = 8300.00

As you can see Bill has more by $85.

^Check yourself

5 0

2 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

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

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

5 0

2 years ago