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3 years ago

Item 11

Mathematics

1 answer:

marysya [2.9K]3 years ago

7 0

Answer:

4 ft higher

Step-by-step explanation:

Since the ladder is 10 ft long and its top is 6 feet high(above the ground), we find the distance of its base from the wall since these three (the ladder, wall and ground) form a right angled triangle. Let d be the distance from the wall to the ladder.

So, by Pythagoras' theorem,

10² = 6² + d² (the length of the ladder is the hypotenuse side)

d² = 10² - 6²

d² = 100 - 36

d² = 64

d = √64

d = 8 ft

Since the ladder is moved so that the base of the ladder travels toward the wall twice the distance that the top of the ladder moves up.

Now, let x be the distance the top of the ladder is moved, the new height of top of the ladder is 6 + x. Since the base moves twice the distance the top of the ladder moves up, the new distance for our base is 8 - 2x(It reduces since it gets closer to the wall).

Now, applying Pythagoras' theorem to the ladder with these new lengths, we have

10² = (6 + x)² + (8 - 2x)²

Expanding the brackets, we have

100 = 36 + 12x + x² + 64 - 32x + 4x²

collecting like terms, we have

100 = 4x² + x² + 12x - 32x + 64 + 36

100 = 5x² - 20x + 100

Subtracting 100 from both sides, we have

100 - 100 = 5x² - 20x + 100 - 100

5x² - 20x = 0

Factorizing, we have

5x(x - 4) = 0

5x = 0 or x - 4 = 0

x = 0 or x = 4

The top of the ladder is thus 4 ft higher

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from the edge of a 486-foot cliff, peyton shot an arrow over the ocean with an initial upward velocity of 90-feet per second

anygoal [31]

Answer:

9 seconds

Step-by-step explanation:

The complete question is

The altitude of an object, d, can be modeled using the equation below:

d=-16t^2 +vt+h

from the edge of a 486 foot cliff, Peyton shot an arrow over the ocean with an initial upward velocity of 90 feet per second. In how many seconds will the arrow reach the water below?

Let

d ----> the altitude of an object in feet

t ---> the time in seconds

v ---> initial velocity in ft per second

h ---> initial height of an object in feet

we have

$d=-16t^2 +vt+h$

we know that

When the arrow reach the water the value of d is equal to zero

we have

$v=90\ ft/sec$

$h=486\ ft$

$d=0\ ft$

substitute the values and solve for t

$0=-16t^2 +(90)t+486$

$-16t^2+90t+486=0$

Multiply by -1 both sides

$16t^2-90t-486=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$a=16\\b=-90\\c=-486$

substitute in the formula

$t=\frac{90(+/-)\sqrt{-90^{2}-4(16)(-486)}}{2(16)}$

$t=\frac{90(+/-)\sqrt{39,204}}{32}$

$t=\frac{90(+/-)198}{32}$

$t_1=\frac{90(+)198}{32}=9\ sec$

$t_2=\frac{90(-)198}{32}=-3.375\ sec$

the solution is t=9 sec

see the attached figure to better understand the problem

4 0

3 years ago

Pls help this is due in 6 minutes I will give u 100 points

vaieri [72.5K]

Answer:

yep

Step-by-step explanation:

4 0

3 years ago

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A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb

Maksim231197 [3]

Answer:

a) $P(X = 1) = 0.38742$

b) $P(X = 3) = 0.05740$

c) $P(X = 9) = 0.00000$

d) $P(X \geq 5) = 0.00163$

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem

There are 10 containers, so $n = 10$ .

A food-packaging apparatus underfills 10% of the containers, so $p = 0.1$ .

a) This is P(X = 1)

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

b) This is P(X = 3)

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

c) This is P(X = 9)

$P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000$

d) This is $P(X \geq 5)$ .

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

$P(X < 5) + P(X \geq 5) = 1$

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868$

$P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742$

$P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937$

$P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740$

$P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837$

Finally

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163$

3 0

3 years ago

Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that

V125BC [204]

Answer:

-1.39

Step-by-step explanation:

Revenue and cost as a function of units sold are $u(x) = 9x-2x^{2}$ and $c(x)=x^{3}-3x^{2}+4x+1$ respectively.

we are have to know for which value or input units are these functions at maximum which translates to for how many units is the revenue maximum and for how many same units is our cost minimum.

5 0

3 years ago

Each side of a cube 2.5 cm long. what is the volume?

Firdavs [7]

V ≈ 15.63cm^3

volume of a cube = side × side × side = units^3

Volume = 2.5*2.5*2.5 = 15.63^3

<em>Hoped this helped!</em>

7 0

3 years ago

Read 2 more answers