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ira [324]
3 years ago
10

PLEASE HELP I WILL BRAINLIST

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
3 0

Answer:

8

Step-by-step explanation:

Since they intersect you can just solve for x and y

since y=-4x and y=x^2+2x+8, you can set -4x=x^2+2x+8

add 4x to both sides to make 0=x^2+6x+8

and factor out (x+4)(x+2)

so the solutions for x are -4 and -2

then insert the two values for x into the first equation

so y=-4(-4) and y=-4(-2)

so y=16 and y=8 when x=-4 and x=-2

so the smallest value for y in both equations is 8

Please give brainliest :)

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The domain is the set of first numbers of the ordered pairs: {-1, 0, 1, 2}
5 0
3 years ago
What is the equation of a line that passes through the points (–3, 4) and (2, 8)?
Mrac [35]

Answer:

\frac{y - 4}{x - ( -  3)}  =  \frac{8 - 4}{2  -  (-  3)}  \\  \frac{y - 4}{x + 3}  =  \frac{4}{5}  \\ 5(y - 4) = 4(x + 3) \\ 5y - 20 = 4x + 12 \\  \boxed{4x - 5y + 32 = 0}

<h2>4x-5y+32=0 is the right answer.</h2>
5 0
3 years ago
A telemarketer earns $150 each week plus $2 for each call that results in a sale.
zepelin [54]

Answer:

P = 2x + 150

Step-by-step explanation:

P = 2x + 150

8 0
3 years ago
You have P = 472 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the r
Ray Of Light [21]
The width would be 236 and the lengths would be 118

Use the equations 2L+W=472 and W*L=MAX
Change the first equation to W=472-2L and plug this into the other equation
(472-2L)(L)=MAX
472L-2L^2=M (take derivative)
472-4L=0 (set to 0 to find the max value)
4L=472 
L=118 
Plug into original to get W=236

Hope this helps!
4 0
3 years ago
Maria y juan deben realizar el mapamundi en medio pliego de cartulina para la tarea de sociales pero cada uno tiene un cuarto de
kondor19780726 [428]

Answer:

Cuando María afirma que si unen sus dos cuartos de cartulina obtendrán el medio pliego que necesitan, esto es:

  • <u>Verdadero</u>.

Step-by-step explanation:

Para entender mejor el ejercicio vamos a utilizar números cada vez que se habla de cantidades de cartulina, por lo tanto, María y Juan tienen 1/4 de cartulina cada uno, es decir, 1/4 * 2, y necesitan 1/2 pliego para poder realizar su tarea, por lo tanto, con la afirmación de María sobre unir los dos cuartos de cartulina, en caso de que sea verdadero, ocurrirá que la suma de los dos cuartos dará el medio pliego, como se muestra a continuación:

  • Total de cartulina de María y Juan = \frac{1}{4}+\frac{1}{4}
  • Total de cartulina de María y Juan = \frac{4+4}{16}
  • Total de cartulina de María y Juan = \frac{8}{16}

Procedemos a simplificar el fraccionario obtenido, sacando mitad tanto en el numerador como en el denominador:

  • Total de cartulina de María y Juan = \frac{4}{8}
  • Total de cartulina de María y Juan = \frac{2}{4}
  • <u><em>Total de cartulina de María y Juan = </em></u>\frac{1}{2}<u><em /></u>

Como puedes ver al final, <u>la cantidad de cartulina de ambos, al ser sumada, da como resultado el 1/2 (medio) pliego que necesitan para su tarea de sociales, por lo cual la afirmación de María es correcta</u>.

3 0
3 years ago
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