MATHEMATICS COLLEGE

Solve for x. x - 1
2 = 4

Answers

Answer 1
Answer:

Answer:

x = 16

Step-by-step explanation:

x - 12 = 4

 + 12 = +12

x = 16
Answer 2
Answer:

Answer:

16

Step-by-step explanation:

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Answers

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4. A sphere with a diameter of 16 mm has the same surface area as the total surface area of a right cylinder with the base diameter equal to the sphere diameter.How high is the cylinder?
O A. 8 mm
OB. 14 mm
O C. 16 mm
O D. 18 mm

Answers

Answer: The answer is A, 8 mm.

Step-by-step explanation:

An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.912 + 20t + 65. What is theobject's maximum height?

Answers

Answer:

The maximum height is the y-value of the vertex.

h(t) = -4.9t² + 20t + 65

     a=-4.9   b=20  c=65

h(2) = -4.9(2)² + 20(2) + 65

     = -19.6 + 40 + 65

     = 85.4

Step-by-step explanation:

Assume that the given procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean mu and standard deviation sigma. ​Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma. In an analysis of preliminary test results from a​ gender-selection method, 33 babies are born and it is assumed that​ 50% of babies are​ girls, so equals 33 and p equals 0.5.

Answers

Answer:

Step-by-step explanation:

Hello!

The study variable is:

X: Number of female babies in a sample of 33 babies.

The variable has binomial distribution, symbolically:

X~Bi(n; p)

n= 33

p= 0.5

The mean of the Binomial variable is:

E(X)= n*p

E(X)= 33*0.5

E(X)= 16.5

The variance of the Binomial is:

V(X)= n*p*q

q= (1 - p) (If "p" is the probability of "success", "q" represents the probability of "failure")

V(X)= 33*0.5*0.5

V(X)= 8.25

Then the standard deviation is:

√V(X)= √8.25= 2.87

E(X) + 2*(√V(X))= 22.24

E(X) - 2*(√V(X))= 10.76

I hope it helps!

Arange the number line in the following spending order 20,-40,5-1,10​

Answers

-40 4 10 20 I think this is the answer I’m not sure

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 meters in t sec.a. Find the​ rock's velocity and acceleration at time t.
b. How long does it take the rock to reach its highest​ point?
c. How high does the rock​ go?
d. How long does it take the rock to reach half its maximum​ height?
e. How long is the rock​ a loft?

Answers

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)=(ds)/(dt)
  • Acceleration is defined as the rate of change of velocity. a(t)=(dv)/(dt)

a.

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

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

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

a(t)=(d)/(dt)(36-1.6t)\n\na(t)=-1.6

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

v(t)=36-1.6t=0\n36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\n360-16t=0\n360-16t-360=0-360\n-16t=-360\nt=(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\ns(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 = (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\n36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\n360t-8t^2=2025\n360t-8t^2-2025=2025-2025\n-8t^2+360t-2025=0

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

x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\n\nt=(-360+√(360^2-4\left(-8\right)\left(-2025\right)))/(2\left(-8\right))=(45\left(2-√(2)\right))/(4)\approx 6.59\n\nt=(-360-√(360^2-4\left(-8\right)\left(-2025\right)))/(2\left(-8\right))=(45\left(2+√(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\n\n\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\n\n\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\n\nt=0,\:t=45

The rock is aloft for 45 seconds.
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