Find the rule and nth term for the sequence 4, 7, 10, 13, 16

The length of the hypotenuse of a right triangle is 24. If the length of one leg is 8, what is approximate length of the other l

Verdich [7]

Something that a right triangle is characterised by is the fact that we may use Pythagoras' theorem to find the length of any one of its sides, given that we know the length of the other two sides. Here, we know the length of the hypotenuse and one other side, therefor we can easily use the theorem to solve for the remaining side.

Now, Pythagoras' Theorem is defined as follows:

c^2 = a^2 + b^2, where c is the length of the hypotenuse and a and b are the lengths of the other two sides.

Given that we know that c = 24 and a = 8, we can find b by substituting c and a into the formula we defined above:

c^2 = a^2 + b^2

24^2 = 8^2 + b^2 (Substitute c = 24 and a = 8)

b^2 = 24^2 - 8^2 (Subtract 8^2 from both sides)

b = √(24^2 - 8^2) (Take the square root of both sides)

b = √512 (Evaluate 24^2 - 8^2)

b = 16√2 (Simplify √512)

= 22.627 (to three decimal places)

I wasn't sure about whether by 'approximate length' you meant for the length to be rounded to a certain number of decimal places or whether you were meant to do more of an estimate based on your knowledge of surds and powers. If you need any more clarification however don't hesitate to comment below.

4 0

3 years ago

12x -20=n (3x-5) I need help I don’t get it please help me.

Salsk061 [2.6K]

Answer:

The value of n = 4 and

The value of x = $\frac{5}{3}$

Step-by-step explanation:

Method I:

As given, 12x -20=n (3x-5) .........(1)

Firstly , take 12x - 20

As 12x - 20 can be written as 4(3x - 5)

∴ equation (1) becomes

4(3x - 5) = n (3x-5)

Divide the equation by (3x-5) , we get

4 = n

Method II:

As given, 12x -20=n (3x-5)

⇒12x - 20 = 3nx - 5n

⇒12x - 3nx = 20 - 5n

⇒3x(4 - n) = 5( 4 - n)

Divide the equation by (4-n) , we get

3x = 5

⇒ x = $\frac{5}{3}$

∴ we get

n = 4 and x = $\frac{5}{3}$

Now, check that value of n and x satisfies the equation or not

12( $\frac{5}{3}$ ) - 20 = 4( 3( $\frac{5}{3}$ ) - 5)

⇒4(5) - 20 = 4(5 - 5)

⇒20 - 20 = 4(0)

⇒ 0 = 0

Hence , we get

The values of x and n satisfies the equation

so, we get

n = 4 and x = $\frac{5}{3}$

5 0

3 years ago

a system of equations is shown below: equation a: 2c = d − 8 equation b: c = 3d 8 which of the following steps should be perform

Fofino [41]

If you would like to know which of the following steps should be performed to eliminate variable d first, you can find this using the following steps:

equation a: 2c = d - 8 /*(-3)
equation b: c = 3d + 8
__________________
-6c = -3d + 24
c = 3d + 8
__________________
-6c + c = -3d + 24 + 3d + 8
-5c = 32
5c = -32
c = -32/5

The correct result would be: multiply equation a by −3.

3 0

4 years ago

An elevator has a placard stating that the maximum capacity is 1600 lb�10 passengers. So, 10 adult male passengers can have a m

Bezzdna [24]

Answer:

1. 0.7421 = 74.21% probability the elevator is overloaded.

2. D.No, there is a good chance that 10 randomly selected people will exceed the elevator capacity.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .