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3 years ago

HALPPPPP Let g(x) = 2x and h(x) = x2 + 4. Evaluate each expression: g(-2) + h(4)

1 answer:

5 0

Answer

16

Explanation

Functions are:

g(x) = 2x and h(x) = x2 + 4

To evaluate g(-2) substitute x = -2 into the function.

2 • -2 = -4

To evaluate h(4) substitute x = 4 into the function.

4^2 + 4 = 20

Add the 2 functions

-4 + 20 = 16

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Cuantos yenes se pueden comprar con 200 pesos mx ? si los yenes cuestan 0.178 centavos ayuda rapido

Sergeeva-Olga [200]

Usando proporciones, hay que con 200 pesos mx se pueden comprar 0.0645 yenes.

<h3>¿Qué es una proporción?</h3>

Una proporción es una fracción de la cantidad total, y puede ser encontrada por intermedio de una regla de tres.

En este problema, hay que un yene cuesta $0.17, enquanto $1 = 3100 pesos. Por eso, cada yene es equivalente a 3100 pesos.

Por eso, la regla de tres es:

1 yene - 3100 pesos

x yenes - 200 pesos

Aplicando multiplicación cruzada:

$3100x = 200$

$x = \frac{200}{3100}$

$x = 0.0645$

Con 200 pesos mx se pueden comprar 0.0645 yenes.

Puede-se aprender más a cerca de proporciones en brainly.com/question/26395519

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3 years ago

Rewrite the expression with grouping symbols 5m+5n

Romashka-Z-Leto [24]

Factorise the 5
5m + 5n = 5(m+n)

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3 years ago

Read 2 more answers

Which values are solutions to the inequality below? Check all that apply. √x >8

lapo4ka [179]

Answer:

x > 64

therefore, 65 and 66 both apply

Step-by-step explanation:

$\sqrt{x}$ > 8

square each side to get:

x > 64

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3 years ago

The number of hours, h, spent studying and the score, s, on a test. What is the dependent variable?

Savatey [412]

Answer: s is the dependent variable

Step-by-step explanation:

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3 years ago

What are the solutions of x^2-2x+5=0

sweet [91]

The solution of $x^{2}-2 x+5=0$ are 1 + 2i and 1 – 2i

Solution:

Given, equation is $x^{2}-2 x+5=0$

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ --- (1)

Let us determine the nature of roots:

Here in $x^{2}-2 x+5=0$ a = 1 ; b = -2 ; c = 5

$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$

Since $b^2 - 4ac < 0$ , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}$

On solving we get,

$x=\frac{2 \pm \sqrt{4-20}}{2}$

$x=\frac{2 \pm \sqrt{-16}}{2}$

$x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}$

we know that square root of -1 is "i" which is a complex number

$\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}$

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

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3 years ago