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ira [324]
2 years ago
5

In the polynomial function below, what is the leading coefficient?

Mathematics
1 answer:
Andrei [34K]2 years ago
5 0

Answer:

search it up see if you could find the answer

Step-by-step explanation:

search it up

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Use the graph to find the slope (m) and y-intercept (b)
irga5000 [103]

Answer:

Slope=4 y-intercept=(0,-21) or -21

Step-by-step explanation:

To find<u> slope</u>

m=-13-(-17)/2-1

m=4/1

m=4

To find <u>y-intercept</u>

y-(-17)=4(x-1)

y+17=4(x-1)

y+17=4x-4

Subtract both sides by 17

y=4x-21

y-intercept=-21

8 0
3 years ago
Find the domain of f(x)=sec(2x)
rjkz [21]

Answer:

*Refer the image attached

Step-by-step explanation:

*Refer the image attached

6 0
3 years ago
In which graph does y vary directly as x? It’s not C, I got it wrong.
Ludmilka [50]
Hello there!

Based on my research and information, I really believe that (option a) would actually be your correct answer. I believe this basically because sense the letter k is always constant, we would only divide this by what it is, which is y, and from these, your solution would always look like your (option a). Please inform if correct.

Happy to help!
4 0
3 years ago
Will give brainliest please help asap
Assoli18 [71]

\sqrt{54}
is same as
3 \sqrt{6}



\sqrt{24}
is same as
2 \sqrt{6}


no you have same sq. root
2 \times 3 \sqrt{6}  + 5 \times 2 \sqrt{6}
6 \sqrt{6}  + 10 \sqrt{6}
16 \sqrt{6}

7 0
3 years ago
Ecuación de la hipérbola con centro en (0;0), focos en abrir paréntesis 0 coma espacio menos raíz cuadrada de 28 cerrar paréntes
yaroslaw [1]

Answer:

\frac{y^{2}}{25}-\frac{x^{2}}{3}=1

Step-by-step explanation:

Para resolver este problema debemos tomar en cuenta los datos que nos dan y la ecuación de una hipérbola. Comencemos con los datos:

centro: (0,0)

focos: (0,-\sqrt{28}),(0,\sqrt{28})

eje conjugado = 2\sqrt{3}

por los focos podemos ver que la hipérbola se dirige hacia el eje y, por lo que debemos tomar la siguiente forma de la ecuación de la parábola:

\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1

de los focos podemos obtener que:

c=\sqrt{28}

y del eje conjugado podemos saber que al dividir la longitud del eje conjugado dentro de 2 obtenemos b, así que:

b=\sqrt{3}

podemos utilizar la siguiente fórmula para obtener a:

c^{2}-a^{2}=b^{2}

si despejamos a en la ecuación obtenemos lo siguiente:

a=\sqrt{c^{2}-b^{2}}

ahora podemos sustituir los valores:

a=\sqrt{(\sqrt{28})^{2}-(\sqrt{3})^{2}}

a=\sqrt{28-3}

a=\sqrt{25}

a=5

así que media vez conozcamos a, podemos sustituir los datos en la ecuación de la hipérbola así que obtenemos lo siguiente:

\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1

\frac{y^{2}}{(5)^{2}}+\frac{x^{2}}{(\sqrt{3})^{2}}=1

\frac{y^{2}}{25}+\frac{x^{2}}{3}=1

si graficamos la hipérbola, queda como en el documento adjunto.

7 0
3 years ago
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