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3 years ago

A die is rolled twice. what is the probability of showing a 1 on the first roll and an even number on the second roll?

Mathematics

1 answer:

marta [7]3 years ago

4 0

The probability of it showing a 1 is 1/6 and the probability of it showing an even number on the second roll is 3/6.

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

DiKsa [7]

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.

6 0

4 years ago

If you begin a problem with a one digit integer , multiply the number by 4 add 7 divide by 3 subtract 5 and answer of 0 what is

12345 [234]

Answer: imma figure this out

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the value of x in the equation 4(2x + 10) = 0? (1 point)

miskamm [114]

Answer:

x = -5

Explanation:

4(2x + 10) = 0

[ Simplify both sides of the equation ]

4(2x + 10) = 0

(4)(2x) + (4)(10) = 0 [Distribute]

8x + 40 = 0

[ Subtract 40 from both sides ]

8x + 40 − 40 = 0 − 40

8x = −40

[ Divide both sides by 8 ]

8x / 8 = −40 / 8

x = -5

4 0

3 years ago

What am I supposed to do?

qaws [65]

its a virus dont fall for it

Step-by-step explanation:

6 0

2 years ago

Match the following with their formula.

neonofarm [45]

For simplicity, I’m marking the six different terms on the main part of the image as 1-6 and the six equations/points on the bottom of the image as A-F.

First, we can look at the bottom and see that there are four equations, but A and F are both points. Looking at the terms above, the only ones that are points are 1 and 6. Since the y-intercept is when a function passes through the y-axis, that means that the x value of the point must be 0. Therefore, A matches with 6.

That leaves F to match with 1. The equation (-b/2a) is what is used in a quadratic equation to locate the vertex, so it makes sense that the vertex would be when you put that number into the equation.

Now on to the four different equations. One of the equations, C, is calculating an x value. This means that it has to match with 2 because it’s drawing a line vertically through the point we already determined to be the vertex of a parabola in the above paragraph.

For the three functions that remain, B has no numbers in it, only the standard letters as placeholders for various numbers you would put into a quadratic function. Therefore, it has to match with 3 since it’s just the standard equation.

That leaves us with D and E to match with 4 and 5, meaning that one of them has to be a negative equation and the other has to be a positive equation. This is where you look at the a value in the function. If the a value is positive, then the quadratic opens up (with the tip pointed down) while if the a value is negative, then the quadratic opens down (with the tip pointed up). This means that D matches with 4 and E matches with 5.

In conclusion,
1 = F
2 = C
3 = B
4 = D
5 = E
6 = A

Hope that helped!

3 0

3 years ago