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2 years ago

A traingle has angles that measure 90° and 50° what is the measure of the third angle?

Mathematics

1 answer:

zvonat [6]2 years ago

7 0

Answer:

We have the following information :

Measure of first angle : $90$ °

Measure of second angle : $50$ °

Measure of third angle :

We know that the angles of a triangle is 180°

$90 + 50 + x =180$

$140 + x = 180$

moving 140 to the other side, $x = 180-140$

$x = 40$ °

Measure of third angle is 40°

Please mark me as brainliest if you found this helpful. Good day ahead.

Dan

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8_murik_8 [283]

Step-by-step explanation:

let m2 be x

then m1 = 4x

Since, m1+m2 = 115

therefore x + 4x = 115

5x = 115

x = 23

so , m2 = 23°

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2 years ago

Read 2 more answers

A gymnast dismounts the uneven parallel bars. Her height, h, depends on the time, t, that she is in the air h = −162 + 24t + 16.

dimulka [17.4K]

That is when h=0
so assuming yo meant
h(t)=-16t²+24t+16
solve for t such that h(t)=0
because when height=0, the gymnast hits the ground

so
0=-16t²+24t+16
using math (imma complete the square
0=-16(t²-3/2t)+16
0=-16(t²-3/2t+9/16-9/16)+16
0=-16((t-3/4)²-9/16)+16
0=-16(t-3/4)²+9+16
0=-16(t-3/4)²+25
-25=-16(t-3/4)²
25/16=(t-3/4)²
sqrt both sides
+/-5/4=t-3/4
3/4+/-5/4=t
if we do plus (because minus would give us negative height)
8/4=t
2=t
it takes 2 seconds

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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

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