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3 years ago

Use a sum identity to find the exact value of cos 285.

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

8 0

The exact value of cos 285 is:
Fraction: (square root of 6 - square root of 2) over 4
Decimal: 0.25881904...

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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

29. The table below represents the relationship of the price of carrots to their weight in ounces in a market: Price Weight, in

Alinara [238K]

The proportional relationship that represents the market that sells their carrots at a lower rate per pound is: y = 1.6x.

<h3>What is a proportional relationship?</h3>

A proportional relationship is a function in which the output variable is given by the input variable multiplied by a constant of proportionality, that is:

y = kx

In which k is the constant of proportionality.

From the table, the rate is:

k = $8.8/4.4 = 2.

From the market, the rate is:

k = $4.8/3 = $1.6.

Hence the equation is y = 1.6x.

More can be learned about proportional relationships at brainly.com/question/10424180

#SPJ1

3 0

2 years ago

I.5y=2x-5<br> II.5y=4+3x<br> III.5y-3x=-1

Ratling [72]

Answer:

Step-by-step explanation:

the vale of x differs in each situation are we talking AB value

6 0

3 years ago

Convert 54,000,000,000 to scientific notation

kvasek [131]

Answer:

5.4 * 10^10

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Evaluate the indefinite integral. <br> integar x4/1 + x^10 dx

ivann1987 [24]

Answer:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c$

Step-by-step explanation:

Given

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

Required

Integrate

We have:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

Let

$u = x^5$

Differentiate

$\frac{du}{dx} = 5x^4$

Make dx the subject

$dx = \frac{du}{5x^4}$

So, we have:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

$\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}$

$\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du$

Express x^(10) as x^(5*2)

$\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du$

Rewrite as:

$\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du$

Recall that: $u = x^5$

$\frac{1}{5} \int\ {\frac{1}{1 + u^2}}} \, du$

Integrate

$\frac{1}{5} * \arctan(u) + c$

Substitute: $u = x^5$

$\frac{1}{5} * \arctan(x^5) + c$

Hence:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c$

7 0

3 years ago