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masha68 [24]
3 years ago
13

3(4x+4) pls quickly:)

Mathematics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

12x+12

Step-by-step explanation:

3(4x+4)

you times both answers to get rid of the 3

=3 • 4x + 3 • 4

= 12x+12

because they are not like terms so you cannot do anything further.

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April shoots an arrow upward into the air at a speed of 32 feet per second from a platform that is 11 feet high. The height of t
dusya [7]

Answer:

27 ft

the maximum height of the​ arrow is 27 ft

Step-by-step explanation:

Given;

The height of the arrow is given by the function;

h(t) = -16t^2 + 32t + 11

Maximum height is at point when dh(t)/dt = 0.

Differentiating h(t), we have;

dh/dt = -32t + 32 = 0

Solving for t;

-32t = -32

t = -32/-32 = 1

t = 1 (time at maximum height is t = 1)

Substituting t=1 into h(t), to determine the value of maximum height;

h(max)= -16(1^2) + 32(1) + 11

h(max) = 27 ft

the maximum height of the​ arrow is 27 ft.

8 0
3 years ago
Ples help me i woud be aprishated
worty [1.4K]

Answer:

Number 2 is  A

Step-by-step explanation:

5 is 380% because 405-22 is 383 if you round that it is 380%

5 0
3 years ago
Read 2 more answers
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
3±i write the quadratic equation with the given roots
Katyanochek1 [597]
By null factor Law factorised eqn=
(x-3+i)(x-3-i)
x^2-6x+10
7 0
4 years ago
4x : 3 = 6 : 5<br> Calculate the value of x.
Goryan [66]

Answer:

x = 9/10

Step-by-step explanation:

This problem features a ratio: 4x/3 = 6/5

By cross multiplying you get that 4x*5 = 3*6 or 20x = 18. By dividing both sides by 20, you get that x = 18/20, and when simplified, 9/10.

5 0
3 years ago
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