All categories

3 years ago

Please help! Both problems are below and please make sure to answer both! Will give brainiest if possible!

Mathematics

2 answers:

valentina_108 [34]3 years ago

3 0

Answer:

picture a is 180⁰ clockwise or counterclockwise (c)

picture b is 90⁰ counterclockwise (b)

Step-by-step explanation:

hope this helps :)

chubhunter [2.5K]3 years ago

3 0

Answer:

A and B

Step-by-step explanation:

The first one is 180 clockwise…

Second picture is 90 counterwise.

I hope thi help and have a good day.

You might be interested in

How do I evaluate 564÷24 with a fraction or decimal

Butoxors [25]

564/24 equals to 47/2 or 23.5

4 0

3 years ago

Answer plzzzz I need it ok

Mariana [72]

Answer:

There are 26 different possible outcomes. the probability of the computer choosing the first letter of your name is 1 out of 26.

4 0

3 years ago

Find the value of the variable.

MatroZZZ [7]

Answer:

sin 30=14/x

x=14*2

x=28

and

x=√{28²-14²}

x=14√3

7 0

3 years ago

Find how many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6 and 7 (with repetitions), if:

Goshia [24]

Answer:

case 1 = 2592

case 2 = 729

case 1 + case 2 = 2916

(this is not a direct adition, because case 1 and case 2 have some shared elements)

Step-by-step explanation:

Case 1)

6 digits numbers that can be divided by 25.

For the first four positions, we can use any of the 6 given numbers.

For the last two positions, we have that the only numbers that can be divided by 25 are numbers that end in 25, 50, 75 or 100.

The only two that we can create with the numbers given are 25 and 75.

So for the fifth position we have 2 options, 2 or 7,

and for the last position we have only one option, 5.

Then the total number of combinations is:

C = 6*6*6*6*2*1 = 2592

case 2)

The even numbers are 2,4 and 6

the odd numbers are 3, 5 and 7.

For the even positions we can only use odd numbers, we have 3 even positions and 3 odd numbers, so the combinations are:

3*3*3

For the odd positions we can only use even numbers, we have 3 even numbers, so the number of combinations is:

3*3*3

we can put those two togheter and get that the total number of combinations is:

C = 3*3*3*3*3*3 = 3^6 = 729

If we want to calculate the combinations togheter, we need to discard the cases where we use 2 in the fourth position and 5 in the sixt position (because those numbers are already counted in case 1) so we have 2 numbers for the fifth position and 2 numbers for the sixt position

Then the number of combinations is

C = 3*3*3*3*2*2 = 324

Case 1 + case 2 = 324 + 2592 = 2916

4 0

3 years ago

La potencia que se obtiene de elevar a un mismo exponente un numero racional y su opuesto es la misma verdadero o falso?

malfutka [58]

Answer:

Falso.

Step-by-step explanation:

Sea $d = \frac{a}{b}$ un número racional, donde $a, b \in \mathbb{R}$ y $b \neq 0$ , su opuesto es un número real $c = -\left(\frac{a}{b} \right)$ . En el caso de elevarse a un exponente dado, hay que comprobar cinco casos:

(a) El exponente es cero.

(b) El exponente es un negativo impar.

(c) El exponente es un negativo par.

(d) El exponente es un positivo impar.

(e) El exponente es un positivo par.

(a) El exponente es cero:

Toda potencia elevada a la cero es igual a uno. En consecuencia, $c = d = 1$ . La proposición es verdadera.

(b) El exponente es un negativo impar:

Considérese las siguientes expresiones:

$d' = d^{-n}$ y $c' = c^{-n}$

Al aplicar las definiciones anteriores y las operaciones del Álgebra de los números reales tenemos el siguiente desarrollo:

$d' = \left(\frac{a}{b} \right)^{-n}$ y $c' = \left[-\left(\frac{a}{b} \right)\right]^{-n}$

$d' = \left(\frac{a}{b} \right)^{(-1)\cdot n}$ y $c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{(-1)\cdot n}$

$d' = \left[\left(\frac{a}{b} \right)^{-1}\right]^{n}$ y $c' = \left[(-1)^{-1}\cdot \left(\frac{a}{b} \right)^{-1}\right]^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c = (-1)^{n}\cdot \left(\frac{b}{a} \right)^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left[(-1)\cdot \left(\frac{b}{a} \right)\right]^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left[-\left(\frac{b}{a} \right)\right]^{n}$

Si $n$ es impar, entonces:

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = - \left(\frac{b}{a} \right)^{n}$

Puesto que $d' \neq c'$ , la proposición es falsa.

(c) El exponente es un negativo par.

Si $n$ es par, entonces:

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left(\frac{b}{a} \right)^{n}$

Puesto que $d' = c'$ , la proposición es verdadera.

(d) El exponente es un positivo impar.

Considérese las siguientes expresiones:

$d' = d^{n}$ y $c' = c^{n}$

$d' = \left(\frac{a}{b}\right)^{n}$ y $c' = \left[-\left(\frac{a}{b} \right)\right]^{n}$

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{n}$

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = (-1)^{n}\cdot \left(\frac{a}{b} \right)^{n}$

Si $n$ es impar, entonces:

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = - \left(\frac{a}{b} \right)^{n}$

(e) El exponente es un positivo par.

Considérese las siguientes expresiones:

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = \left(\frac{a}{b} \right)^{n}$

Si $n$ es par, entonces $d' = c'$ y la proposición es verdadera.

Por tanto, se concluye que es falso que toda potencia que se obtiene de elevar a un mismo exponente un número racional y su opuesto es la misma.

3 0

3 years ago