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3 years ago

In a sequence of numbers, a4=12, a5=14, a6=16, a7=18, and a8=20.

Mathematics

1 answer:

faltersainse [42]3 years ago

6 0

Answer:

you are right, it's a)❤

Step-by-step explanation:

a4=12---> put the formula in it

2×4+4=12

a5=14---> 2×5+4=14

a6=16---> 2×6+4=16

hope it helps❤

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I need help finding x!<br><br> Please show work for both!

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1. Answer x=-5
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3 years ago

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Which set of numbers could represent the lengths of the sides of a right triangle?

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Answer:

The first set: 8, 15, and 17.

Step-by-step explanation:

By the pythagorean theorem, a triangle is a right triangle if and only if

$\text{longest side}^2 = \text{first shorter side}^2 + \text{second shorter side}^2$ .

In this case,

$\text{longest side}^2 = 17^2 = 289$ .

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 8^2 + 15^2\\ &=64 + 225 = 289 \end{aligned}$ .

In other words, indeed $\text{hypotenuse}^2 = \text{first leg}^2 + \text{second leg}^2$ . Hence, 8, 15, 17 does form a right triangle.

Similarly, check the other pairs. Keep in mind that the square of the longest side should be equal to the sum of the square of the two

Factor out the common factor $2$ to simplify the calculations.

$\text{longest side}^2 = 20^2 = 400$

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 10^2 + 15^2\\ &=100 + 225 = 325 \end{aligned}$ .

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

$\text{longest side}^2 = (2\times 11)^2 = 2^2 \times 121$ .

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= (2 \times 6)^2 + (2 \times 9)^2\\ &=2^2 \times(36 + 81) = 2^2 \times 117 \end{aligned}$ .

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

$\text{longest side}^2 = 11^2 = 121$ .

<h3> $\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 7^2 + 9^2\\ &=49+ 81 = 130 \end{aligned}$ .</h3>

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

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4 years ago

All of the following are true of betas and correlation coefficients EXCEPT: Group of answer choices Both betas and correlation c

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Answer:

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3 years ago

A certain three-cylinder combination lock has 60 numbers on it. To open it, you turn to a number on the first cylinder, then

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Answer:

(a) The number of different lock combinations is 216,000.

(b) The probability of getting the correct combination in the first try is $\frac{1}{216000}$ .

Step-by-step explanation:

The lock has three-cylinder combinations with 60 numbers on each cylinder.

The procedure of the opening lock is to turn to a number on the first cylinder, then to a second number on the second cylinder, and then to a third number on the third cylinder.

The numbers can be repeated.

(a)

There are three cylinder combinations on the lock, each with 60 numbers.

It is provided that repetitions are allowed.

Then each cylinder can take any of the 60 numbers.

So there are 60 options for the first cylinder.

There are 60 options for the second cylinder.

And there are 60 options for the third cylinder.

So the total number of possible combinations is:

Total number of combinations = 60 × 60 × 60 = 216000.

Thus, the number of different lock combinations is 216,000.

(b)

The events of getting a correct combinations of the three-cylinder combination lock implies that all the three cylinder are set at the correct numbers.

Each cylinder has 60 numbers.

This implies that there are 60 possible ways to get a correct number for the first cylinder.

The probability of getting the correct number for the first cylinder is:

P (Number on 1st cylinder is correct) = $\frac{1}{60}$ .

Similarly for the second cylinder the probability of getting the correct number is:

P (Number on 2nd cylinder is correct) = $\frac{1}{60}$ .

And similarly for the third cylinder the probability of getting the correct number is:

P (Number on 3rd cylinder is correct) = $\frac{1}{60}$ .

So the probability of getting the correct combination in the first try is:

P (Correct combination in the 1st try) = $\frac{1}{60}\times \frac{1}{60}\times \frac{1}{60}=\frac{1}{216000}$ .

Thus, the probability of getting the correct combination in the first try is $\frac{1}{216000}$ .

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3 years ago