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Montano1993 [528]
4 years ago
5

In a 3 hour examination of 350 questions, there are 50 mathematics problems. if twice as much time should be allowed for each ma

thematics problem as for each of the other questions, how many minutes should be spent on the mathematics problems?
Mathematics
1 answer:
Margarita [4]4 years ago
3 0

Answer: 7 minutes

Step-by-step explanation:

350/50=7

3*2.3333=7

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The spinner is 3 times more likely to land on white.

Step-by-step explanation:

The spinner landed 57 times on white and 19 times on black.

57 is 3 * 19

Answer: The spinner is 3 times more likely to land on white.

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Find the 8th term for the sequence a(n)=6-1n
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-2

Step-by-step explanation:

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3 years ago
A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met
Verdich [7]

Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

5 0
4 years ago
Find the composition of
Rus_ich [418]

Answer:

[(x + 6), (y + 1)]

Step-by-step explanation:

Vertices of the quadrilateral ABCD are,

A → (-5, 2)

B → (-3, 4)

C → (-2, 4)

D → (-1, 2)

By reflecting the given quadrilateral ABCD across x-axis to form the image quadrilateral A'B'C'D',

Rule for the reflection of a point across x-axis is,

(x, y) → (x , -y)

Coordinates of the image point A' will be,

A(-5, 2) → A'(-5, -2)

From the picture attached, point E is obtained by translation of point A'.

Rule for the translation of a point by h units right and k units up,

A'(x+h, y+k) → E(x', y')

By this rule,

A'(-5 + h, -2 + k) → E(1, -1)

By comparing coordinates of A' and E,

-5 + h = 1

h = 6

-2 + k = -1

k = 1

That means

Rule for the translation will be,

[(x + 6), (y + 1)]

8 0
2 years ago
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