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1 year ago

Which graph represents f(x)=−2cos(4πx)?

Mathematics

1 answer:

algol [13]1 year ago

3 0

Answer:

the last one (bottom right)

Step-by-step explanation:

4×pi is the same as twice around the whole circle to the beginning. it is the same as 0°, and cos(0) = 1, and then -2×cos(0) = -2

so, we only need to look, where the function reaches -2.

it should be at x = 0, x = 0.5 and x = 1.

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A construction worker drops a hammer while building the Grand Canyon skywalk 4,000ft above the Colorado River. Use the formula t

yan [13]

Answer:

15.8

Step-by-step explanation:

We are asked to find how many seconds it takes for an object to fall from a height of 4,000ft. We will use the given formula, t=h−−√4. Substituting h=4,000, we find

t=4000−−−−√4=15.8

So, the hammer reaches the river after 15.8 seconds.

7 0

2 years ago

When a border of uniform width is placed around a 9 x 12 m

zhenek [66]

Answer:

1080000

Step-by-step explanation:

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2 years ago

Write an equation to represent the problem. Then solve the equation. Two years of local Internet service costs $685, including t

iren2701 [21]

Answer:

(Total - Fee) / Months = $Price per Month

(685 - 85) / 24 = 25

Step-by-step explanation:

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2 years ago

Rewrite the expression using grouping symbols 13x - 29x

zhenek [66]

Does it mean factor it out??
x (13-29)

3 0

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