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3 years ago

Find the value of d

Mathematics

1 answer:

tamaranim1 [39]3 years ago

4 0

C : B = D : E
6 : 5 = D : 3
D = 3.6

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M AEB = 37°, what is m CED?

Oksana_A [137]

Answer: 53 degree

Explanation:

AEB + BEC + CED = 180 degree
37 + 90 + CED = 180
CED = 180 - 127
CED = 53 degree

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3 years ago

Find the missing side of each triangle. leave your answers in simplest radical form

dedylja [7]

X^2 = 4^2 - (2√3)^2
x^2 = 16 - 12
x^2 = 4
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answer
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3 years ago

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago

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DochEvi [55]

182.5 US dollars so yes

the answer is True

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3 years ago

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zubka84 [21]

<span>$84 to 12%
</span>10.08
is the answer

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3 years ago