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Jobisdone [24]
3 years ago
10

A Rosa le gusta jugar con su primo Eduardo utilizando números. Rosa le planteó encontrar dos números que sumados den 15 y que el

doble de uno de ellos sea igual al otro más 3 unidades, ¿De qué números se trata?
Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

Los números son 6 y 9

Step-by-step explanation:

Este problema se puede resolver por medio de un sistema de ecuaciones.

El primer número será x y el segundo número será y.

Sabemos que los dos números suman 15, por lo tanto esto se puede escribir como:

x+y=15

Por otro lado sabemos que el doble de uno de ellos es igual al otro más 3 unidades, esto lo podemos escribir de la siguiente manera:

2x=y+3 (el doble del primero es igual al segundo más 3)

Reescribiendo esta segunda ecuación tenemos:

2x-y=3

Por lo tanto, nuestras dos ecuaciones son:

x+y=15\\2x-y=3

Resolviendo el sistema por el método de reducción observamos que, si sumamos ambas ecuaciones, las y se cancelan y quedamos con:

3x=18\\x=6

Ahora, sustituimos este valor en la primera ecuación para obtener el valor de y:

x+y=15\\6+y=15\\y=15-6\\y=9

Por lo tanto, los números son 6 y 9

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-2=g-9<br><br> Please Help!!!
Sergio039 [100]

Answer:

g = 7

Step-by-step explanation:

you add 9 to 9 and cancle it out . then you add 9 to -2 and you get 7 = g . just switch it around and you get your answer .

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4 years ago
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Seven-eighths minus five-twelfths
Elenna [48]
The answer to your question is A, 11/24





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3 years ago
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consider the circle x^2+y^2-6x-8y=0. a.) find an equation of the tangent line to the circle at the point (0,0). b.) find the cen
inna [77]

The slope of the tangent line of the circle  x^2+y^2-6x-8y=0 is \frac{dy}{dx}:

to find it we use implicit differentiation:

\frac{dy}{dx}(x^2)+\frac{dy}{dx}(y^2)-\frac{dy}{dx}(6x)-\frac{dy}{dx}(8y)=0\\\\2x+2y \cdot\frac{dy}{dx}-6-8\frac{dy}{dx}=0\\\\\frac{dy}{dx}(2y-8)=-2x+6\\\\\frac{dy}{dx}= \frac{-2x+6}{2y-8}= \frac{-x+3}{y-4}

thus the slope of the tangent line at a point (x, y) of the circle is:

m= \frac{dy}{dx}=\frac{-x+3}{y-4}


part a:

m at (0, 0) is (-0+3)/(0-4)=3/(-4)=-3/4

the equation of the tangent line is

(y-0)=(-3/4)(x-0)

y=(-3/4)x


part b)

The equation of the circle can be written in standard form by completing the square:

x^2+y^2-6x-8y=0\\\\x^2-6x+y^2-8y=0\\\\(x^2-6x+9)-9+(y^2-8y+16)-16=0\\\\(x-3)^2+(y-4)^2=5^2


thus the circle has radius (3, 4) and radius 5.


part c.

m=\frac{-x+3}{y-4}=\frac{-6+3}{0-4}= \frac{-3}{-4}= \frac{3}{4}

the equation of the line is:

y-0=(3/4)(x-6)

y=(3/4)x-9/2


d) the lines are y=(-3/4)x   and   y=(3/4)x-9/2

they meet at x:

(-3/4)x=(3/4)x-9/2

(-6/4)x=-9/2

(6/4)x=9/2

(2/2)x=3/1

x=3, 

at x=3, y=(-3/4)x=(-3/4)*3=-9/4


Check the graph, generated using desmos.com

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3 years ago
What is the equation of the line that passes through the point 7,-3 and has a slope of -5
Oksana_A [137]
So i’m not sure about this but you would do the y=mx+b form and then since slope = m it would be y=-5x+b and then you would plug 7 and -3 for y and x and then you would get 7=-5-3=b and then solve for b
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OleMash [197]
1} 2.675 (6) =0.1 or the tenth

2} 804.721 (1) = 0.001 or the thousandth
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4 years ago
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