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2 years ago

Which polynomial function has a leading coefficient of 2, root -4 with multiplicity 3, and root 10 with multiplicity 1?

Mathematics

1 answer:

lana [24]2 years ago

7 0

Answer:

f(x) = 2(x+4)(x+4)(x+4)(x-10)

Step-by-step explanation:

Here, we want to write a polynomial function with leading coefficient of 2, root - 4 with multiplicity 3 ( meaning it appears 3 times);

and root 10 with multiplicity 1

If x = -4

Then in linear form, we have x + 4

if x = 10, in linear form, we have x -10

So the polynomial is;

f(x) = 2(x+4)(x+4)(x+4)(x-10)

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In the game of roulette, a wheel consists of 54 slots numbered 00, 0, 1, 2,..., 54. To play the game, a metal ball is spun a

Gnesinka [82]

Step-by-step explanation:

Consider the provided information.

A wheel consists of 54 slots numbered 00, 0, 1, 2,..., 54

Part A:

The sample space is the slots on which the ball can fall.

Here wheel consists of 54 slots numbered. Thus the sample. So, there are total 56 slots and the sample space is {00, 0, 1, 2,..., 54}

Part B:

There are 56 slots and only one slot marked by 7. There is an equivalent likelihood that it will fall into any of these slots.

Thus, the probability of the metal ball falls into the slot marked 7 is:

P(A) = 1/56 = 0.0178

Hence, the probability of the metal ball falls into the slot marked 7 is 0.0178.

Part C:

The number of odd slots are 1, 3, 5...55.

There are total 27 odd numbers and 56 slots.

Thus, the probability of the ball lands in an “odd” slot is:

P(A) = 27/56 = 0.4821

Hence, the probability of the metal ball falls an “odd” slot is 0.4821.

Part D:

Interpret this probability: 27/56 = 0.4821 …

The probability of the ball will fall on odd slot is around 48.21% or we can say, if we spins the wheel 100 times then ball will land on odd place around 48 times.

5 0

3 years ago

A pebble is dropped into a deep well and 3s later the sound of a splash is heard as the pebble reaches the bottom of the well. T

ankoles [38]

Answer:

$2.89\ \text{s}$

0.11 seconds

$37.4\ \text{m}$

Step-by-step explanation:

$t_1$ = Time taken by the pebble to hit the water

$t_2$ = Time taken by the sound to travel to the observer

$t_1+t_2$ = Time taken by the observer to listen to the splash after stone is thrown = 3 s

$t_1=3-t_2$

a = g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$u_s$ = Speed of sound = 340 m/s

The distance traveled by the pebble

$s=ut+\dfrac{1}{2}at_1^2\\\Rightarrow s=0+\dfrac{1}{2}\times9.81(3-t_2)^2\\\Rightarrow s=4.905(9+t_2^2-6t_2^2)$

Distance traveled by the sound

$s=u_st\\\Rightarrow s=340\times t_2$

The distance traveled by the sound and the pebble is equal

$4.905(9+t_2^2-6t_2)=340\times t_2\\\Rightarrow 9+t_2^2-6t_2=\dfrac{340}{4.905}t_2\\\Rightarrow 9+t_2^2-6t_2=69.32t_2\\\Rightarrow t_2^2-75.32t_2+9=0\\\Rightarrow t_2=\frac{-\left(-75.32\right)\pm \sqrt{\left(-75.32\right)^2-4\times \:1\times \:9}}{2\times \:1}\\\Rightarrow t_2=75.2,0.11$