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4 years ago

Why does a y-intercept not count as a zero?

Mathematics

1 answer:

Genrish500 [490]4 years ago

7 0

Answer:

It only counts as a zero when the y-intercept is (0,0).

Step-by-step explanation:

The zeros of a quadratic function are always written as (x,0), while the y-intercept is always written as (0,y). Therefore, in order for a y-intercept to be a zero, it must be (0,0), because the y-coordinate in any zero is 0. At any other time, the y-intercept is not a zero.

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Q(3x)=12x+2 then what is Q(x-2)

Mandarinka [93]

Answer:

it's ax+b form answer is Q(x-2)=4x-6

Step-by-step explanation:

ax+b=Q(x),Q(x-2)=a(x-2)+2

Q(3x)=a(3x)+b=12x+2, 3a=12 =>a=4,b=2

the original q of x is 4x+2

Q of x-2 =4(x-2)+2=4x-8+2=4x-6

8 0

2 years ago

Please please Help me

konstantin123 [22]

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4 0

3 years ago

Write an equation for the circle with endpoints of a diameter (9, 4) and (-3, -2)

inysia [295]

If the diameter is at (9,4) and (-3,-2), then the center of the circle is the midpoint of that segment.

and since we know that the radius of a circle is half of the diameter, whatever long that diameter segment is, the radius is half that.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 9 &,& 4~) 
% (c,d)
&&(~ -3 &,& -2~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-3+9}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \stackrel{center}{(3~,~1)}$

$\bf -------------------------------\\\\
~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 9 &,& 4~) 
% (c,d)
&&(~ -3 &,& -2~)
\end{array}
\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{(-3-9)^2+(-2-4)^2}\implies d=\sqrt{(-12)^2+(-6)^2}
\\\\\\
d=\sqrt{180}\qquad \qquad \qquad radius=\cfrac{\sqrt{180}}{2}$

$\bf -------------------------------\\\\
\textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{3}{ h},\stackrel{1}{ k})\qquad \qquad 
radius=\stackrel{\frac{\sqrt{180}}{2}}{ r}
\\\\\\
(x-3)^2+(y-1)^2=\left( \frac{\sqrt{180}}{2} \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(\sqrt{180})^2}{2^2}
\\\\\\
(x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45$

7 0

3 years ago

Please help with question 6

Kitty [74]

Answer:

y = -3x - 1

Use the methods on your other questions.

6 0

3 years ago

What are the solutions to x2 + x - 5≤0

Juliette [100K]

I have no idea what that x is doing, so imma take it out.

2+x-5≤0

-3+x≤0

+3 +3

x≤3

---

hope it helps

8 0

3 years ago