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3 years ago

In ΔABC, the lengths of a, b, and c are 22.5 centimeters, 18 centimeters, and 13.6 centimeters, respectively.

Mathematics

2 answers:

ehidna [41]3 years ago

8 0

Answer:

It is not 37.19 and 53.13.

Step-by-step explanation:

For newcomers who are using PLATO, I recently finished the test, and it is not the answer below. I don't know the right answer, unfortunately.

Irina-Kira [14]3 years ago

7 0

Given the values of the three sides of the triangle, we can apply the Cosine Law to find the angles of the triangle. Recall that for we can express the value of c through the equation below.

$c^{2} = a^{2} + b^{2} - 2abcosC$

Rearranging this equation, we can find the value ∠C as shown below.

$\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}$
$C = cos^{-1} (\frac{a^{2}+b^{2}-c^{2}}{2ab})$

We can apply the same reasoning for finding the value of ∠B as shown.

$B = cos^{-1} (\frac{a^{2}+c^{2}-b^{2}}{2ac})$

Plugging in the values of the sides (see image attached) from the given. It will now be straightforward to compute for ∠B and ∠C.

$C = cos^{-1} (\frac{22.5^{2}+18^{2}-13.6^{2}}{2(22.5)(18)})$
$C \approx 37.19$

$B = cos^{-1} (\frac{22.5^{2}+13.6^{2}-18^{2}}{2(22.5)(13.6)})$
$B \approx 53.13$

Answer: ∠C = 37.19° and ∠B = 53.13°

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Answer:

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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.

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2 years ago

Rod is saving Php2000 in a bank at the end of each month which gives an interest of 1% compounded monthly.How much is the saving

agasfer [191]

Answer:

Php2040.38

Step-by-step explanation:

Given

$P = 2000$ --- Principal

$r = 1\%$ --- Rate

$t = 2\ years$ --- Time

$n = 12$ --- monthly

Required

Determine the amount at the end of two years

This is calculated as:

$A = P(1 + \frac{r}{n})^{nt}$

So, we have:

$A = 2000(1 + \frac{1\%}{12})^{12*2}$

$A = 2000(1 + \frac{1}{100*12})^{24}$

$A = 2000(1 + \frac{1}{1200})^{24}$

$A = 2000(\frac{1200+1}{1200})^{24}$

$A = 2000(\frac{1201}{1200})^{24}$

$A = 2000*1.02019$

$A = 2040.38$

Hence, the final amount is: Php2040.38

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2 years ago

A motorcycle can travel 60 miles per gallon. Approximately how many gallons of fuel will the motorcycle need to travel 40 km?

Semenov [28]

The first thing to note is since one gallon= 60 miles then 1/2 gallon= 30 miles, now I know 1/2 is not an answer but it is easiest to decrease in moderate amounts. 1/2 gallon= 48 Km so we are close, and 1/2=0.5 so if we brought it down to 0.42 we would have 25.2 miles and 25.2 miles = 40.32. The question wants to know the approximate number and 40.32 is the closest we are going to get so the answer is 0.42

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3 years ago

7. Find the length of HP. 12 P 9 H

bearhunter [10]

Step-by-step explanation:

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3 years ago

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