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3 years ago

Numbers 38 and 40 pls and thanks

Mathematics

2 answers:

Evgen [1.6K]3 years ago

8 0

#40 is $110.16 total. You have to add the sales tax to the amount of the baseball cap which comes to $13.77. Then take that amount and multiply by the number of caps. Hope this helps.

Alekssandra [29.7K]3 years ago

5 0

38a) $10.50 To get this answer, you need to multiply $1.50 by 7

38b) $25.50 To get this answer, you need to subtract $10.50 from $36

40) The answer would be $13.7694 and it could be rounded to $13.77. To get this answer .you need to multiply $12.99 by 1.06 since it has a 6% sales tax.

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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

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A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

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3 years ago

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Step-by-step explanation:

Based on the information given in the question, the equation that can be used to determine the the number of gigabytes of data Peyton can use while staying within her budget will be:

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3 years ago

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You bicycle along a straight flat road with a safety light attached to one foot. Your bike moves at a speed of 15 km/hr and your

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Answer:

x(t) = ( 4.167 t + 0.2 cos (2πt))

y(t) = ( 0.3 - 0.2 sin (2πt))

b) the foot have to be 3.32 rev/sec faster

Step-by-step explanation:

Given that:

the speed of the bike = 15 km/hr = 15 × 1000/3600 (m/sec) = 4.167 m/sec

radius of the circle when the foot moves = 20 cm = 0.2 m

radius of the circle above the ground = 30 cm = 0.3 m

Let assume that:

x(t) should represent the vector along the horizontal moment

y(t) should be the vector along the vertical moment

The initial component will be ( 0, 0.3)

We know that the radius of the circle is given as 0.2 m, So the vector of the circle can be written as (0.2 cos t , 0.2 sin t )

Also, the foot makes one revolution in a second, definitely the frequency of the revolution = 1 and the vector for the circle is ( 0.2 cos (2πt), -0.2 sin (2πt)), due to the fact that the foot moves clockwise.

Thus, adding all the component together ; we have:

(x(t), y(t)) = (0,0.3)+(4.167 t , 0)+(0.2 cos (2πt), -0.2 sin (2πt))

(x(t), y(t)) = (4.167 t + 0.2 cos (2πt), 0.3 - 0.2 sin (2πt))

Hence; the parametric equations are:

x(t) = ( 4.167 t + 0.2 cos (2πt))

y(t) = ( 0.3 - 0.2 sin (2πt))

The linear speed of rotation is :

15km/hr = 15 × 100, 000/3600 (cm/sec)

= 416.7 cm/sec

The rotational frequency is :

= 416.7/2πr

= 416.7/2(3.14 × 20)

= 3.32 rev/sec

Hence, the foot have to be 3.32 rev/sec faster in rotating if an observer standing at the side of the road sees the light moving backward.

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