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sineoko [7]
3 years ago
6

Graph the line by plotting any two ordered pairs that satisfy the equation.

Mathematics
1 answer:
ollegr [7]3 years ago
6 0

Answer:

(0,-4)

(3,0)

Step-by-step explanation:

Let start at the orgin.

This is a linear equation since the equation is in the form of

y = mx + b

where m is the slope and b is the y intercept.

Since we starting at the orgin, and b is our y intercept.

Our first point is

(0,-4).

since the slope is 4/3.

We would rise 4 from the y value and run 3 to the x value.

In other words, to find your second point, go up 4 units from the first point and move to the right 3 units.

So our next point is at

(3,0).

U can continously go up 4 units and move 3 units to the right to find other points.

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The slope of the line below is -0.25. Write the equation of the line in point-
Doss [256]
ANSWER:
y=-0.25x+4.75
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Assuming that boy and girl babies are equally​ likely, what would be​ Kathy's probability of having if she were to have four chi
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It would be 1/2 of a chance
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2 years ago
What is the value of 1/3 x^2 + 5.2y when x = 3 and y = 2?<br><br> PLZ HELP!!!
olasank [31]
The first step in solving this is to substitute x with 3 and y with 2, so that
\frac{1}{3} x^{2} +5.2y
becomes
\frac{1}{3}  (3)^{2} +5.2(2)
Then simply solve using the order of operations.
First exponents (there are no parentheses)
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Then multiply and divide
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And finally add and subtract
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Therefore the value <span>of 1/3 x^2 + 5.2y when x=3 and y=2 is 13.4</span>
6 0
3 years ago
Paula salió de su casa rumbo al trabajo avanzo 8km al este y 13km al norte. Si hubiera un camino recto desde su casa a su trabaj
zhannawk [14.2K]

Answer:

15.26km

Step-by-step explanation:

Con la informacion de la pregunta e hecho un diagrama de la situacion. Podemos ver en el diagrama que el recorido forma un triángulo rectángulo. Entonces para saber la distancia que recorreria por el camino recto desde su casa a su trabajo tendriamos que calcular el valor de x, que se puede calcular usando el teorema de pitagoras que es el siguiente.

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

donde a y b son el ancho y el alto del triangulo y el c seria x

8^{2} + 13^{2}  = x^{2}

64 + 169 = x^{2}

233 = x^{2}

15.26 = x

Finalmente, podemos ver que la distancia seria 15.26km

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%
andrey2020 [161]

First observe that if a+b>0,

(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}

Let a=0 and b=x. It follows that

a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}

Now let b=1, so a^2+a=4x. Solving for a,

a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2

which means

a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}

Now solve for x.

x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2

(note that we assume 2x-1\ge0)

4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}

(we omit x=0 since 2\cdot0-1=-1\ge0 is not true)

3 0
1 year ago
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