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3 years ago

How many pairs of points are reflections across the x-axis? A)1 B)2 C)3 D)4

Mathematics

1 answer:

asambeis [7]3 years ago

6 0

Answer:

C. 3

Step-by-step explanation:

C&D, A&F, K&J

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How does ordering the events this does not affect the level of suspense in the story

vesna_86 [32]

<span>A)The author builds events slowly to create tension.</span>

8 0

3 years ago

Un padre para estimular a su hijo para estudiar matemáticas le dice: por cada ejercicio que resuelvas bien te daré 7 € y por cad

Naya [18.7K]

Answer:

8 ejercicios

Step-by-step explanation:

De acuerdo a la información dada, puedes plantear las siguientes ecuaciones:

x+y= 45 (1)

7x-1y= 19 (2)

x= ejercicios que resolvió bien

y= ejercicios que resolvió mal

Primero, puedes despejar x en (1):

x= 45-y (3)

Segundo, debes reemplazar (3) en (2):

7(45-y)-1y=19

315-7y-1y= 19

315-8y= 19

315-19= 8y

296= 8y

y= 296/8

y= 37

Tercero, debes reemplazar el valor de y en (3) para encontrar x:

x= 45-37

x= 8

De acuerdo a esto, la respuesta es que el hijo ha resuelto bien 8 ejercicios.

8 0

3 years ago

2^X = 17^X<br><br> How do you solve that equation?

Kobotan [32]

For this case we must solve the following equation:

$2 ^ x = 17 ^ x$

For this, we follow the steps below:

We find the Neperian logarithm on both sides of the equation to remove the variable from the exponent

$ln (2 ^ x) = ln (17 ^ x)$

We use one of the logarithm properties to extract x from the exponent:

$xln (2) = xln (7)$

We subtract xln (7) on both sides of the equation:

$xln (2) -xln (7) = 0$

We take x common factor:

$x (ln (2) -ln (7)) = 0$

We divide between $(ln (2) -ln (7))$ on both sides of the equation, then:

$x = 0$

Answer:

$x = 0$

5 0

3 years ago

Which of the following would best be solved using factoring the difference of squares? 2² + 3x - 10 = 0 O 3r² + 12x = 8 2³ +52²

IgorC [24]

Given: Different equations

To Determine: Which would be best solved using difference of two squares

Solution

The factorization of a difference of two squares is given below

$a^2-b^2=(a-b)(a+b)$

Let us examine each of the given equation

$\begin{gathered} x^2+3x-10=0 \\ x^2-2x+5x-10=0 \\ x(x-2)+5(x-2)=0 \\ (x-2)(x+5)=0 \end{gathered}$ $\begin{gathered} 3x^2+12x=8 \\ 3x^2+12x-8=0 \end{gathered}$ $\begin{gathered} x^3+5x^2-9x-45=0 \\ x^2(x+5)-9(x+5)=0 \\ (x+5)(x^2-9)=0 \\ x^2-9=x^2-3^2=(x+3)(x-3) \\ Therefore \\ (x+5)(x+3)(x-3)=0 \end{gathered}$ $\begin{gathered} x^2-25=0 \\ x^2-5^2=0 \\ (x+5)(x-5)=0 \end{gathered}$

From the above,

$\begin{gathered} x^3+5x^2-9x-45=0 \\ AND \\ x^2-25=0 \end{gathered}$

The two equation above can be solved by difference of two square, but the equation below is the easiest solved using differnce of two square

x² - 25 = 0

3 0

1 year ago

Write two numbers that multiply to the value on top and add to the value on the bottom.

sattari [20]

Answer:

6 and 2

Step-by-step explanation:

6 times 2 is 12 and 6 plus 2 is 8

4 0

3 years ago