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dmitriy555 [2]
3 years ago
13

Geometry isosceles and equilateral triangles find the value of x in each diagram

Mathematics
1 answer:
zavuch27 [327]3 years ago
6 0

Answer:

Part 9) x=12

Part 10) x=12\°

Part 11) x=10\°

Part 12) x=11\°

Part 13)

a) SU=4\ units

b) m∠VWX=40\°

c) m∠WVX=50\°

d) m∠XTV=60\°

e) m∠XVT=30\°

Step-by-step explanation:

Part 9) we have that

12=2x-12 ----->by SSA

solve for x

2x=24

x=12

Part 10)

In the isosceles triangle of the left the vertex angle is equal to

180\°-68\°*2=44\°

Find the measure of angle 2

m∠2=90\°-44\°=46\°

m∠2=4x-2

4x-2=46\°

solve for x

4x=48\°

x=12\°

Part 11)

Find the base angle in the isosceles triangle of the top

180\°-118\°=62\°

Find the vertex angle in the isosceles triangle of the top

180\°-2*62\°=56\°

Find the vertex angle 2 in the isosceles triangle of the bottom

180\°-56\°=124\° ------> this is the measure of angle 2

m∠2=124\°

m∠2=12x+4

12x+4=124\°

12x=120\°

x=10\°

Part 12)

m∠2=146\° ------> by corresponding angles

m∠2=13x+3

13x+3=146\°

13x=143\°

x=11\°

Part 13)

a) we have that

SU=UW ------> given problem

3x+1=x+3

2x=2

x=1

therefore

SU=3x+1=3+1=4

SU=4\ units

b) we know that

m∠VWX=4y\°

in the right triangle UVW find the value of y

The sum of the internal angles of a triangle is equal to 180\°

so

180\°=4y+90\°+50\°

180\°=4y+140\°

4y=40\°

y=10\°

so

m∠VWX=4y\°=40\°

Part c) we know that

in the right triangle VWX

The sum of the internal angles of a triangle is equal to 180\°

so

m∠WVX=180\°-(4y+90\°)

m∠WVX=180\°-(40\°+90\°)

m∠WVX=50\°

Part d)  we know that

in the right triangle XTV

The sum of the internal angles of a triangle is equal to 180\°

so

m∠XTV=180\°-(40\°-y+90\°)

m∠XTV=180\°-(40\°-10\°+90\°)

m∠XTV=60\°

Part e) we know that

m∠XVT=(40\°-y)

substitute the value of y

m∠XVT=(40\°-10\°)

m∠XVT=30\°

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interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm
Igoryamba

Answer:

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The normal component of acceleration is a_{\boldsymbol{N}}=1 (2)^2=4

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

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|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2

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