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adelina 88 [10]
3 years ago
11

Please help i'm not that good at math. Determine the number of real solutions for each system of equations.

Mathematics
2 answers:
ratelena [41]3 years ago
7 0

System A has ONE real solution

System B has NO real solutions

System C has ONE real solution

timurjin [86]3 years ago
5 0

A is a circle around the origin and a line through the origin, so they'll intersect in two places; we can skip the detailed calculation.

B is the intersection of a parabola and a line; they can intersect in zero, one or two places depending on the details. Let's check

y=x^2-7x+10=-6x+5

x^2-x+5= 0

Discriminant,

d = b^2-4ac=1-(4)(5)=-19

Negative discriminant, zero real solutions for B.

System C is again a parabola and a line; we proceed similarly:

y = 8x+17 = -2x^2+9

2x^2+8x+8=0

x^2 + 4x + 4 = 0

d=b^2-4ac=16 - 4(1)(4)=0

Zero discriminant, exactly one real solution for C.

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Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).
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So hmm check the picture below

we know the vertex is at the origin, and the focus point is below it, that means two things, the parabola is vertical and it's opening downwards

notice the distance "p", from the vertex to the focus point, is just 7 units, however, since the parabola is opening downwards, the "p" value will be negative, so p = -7

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{4{{ p}}(y-{{ k}})=(x-{{ h}})^2 }\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
4{{ p}}(y-{{ k}})=(x-{{ h}})^2\quad 
\begin{cases}
h=0\\
k=0\\
p=-7
\end{cases}\implies 4(-7)(y-0)=(x-0)^2
\\\\\\
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Step-by-step explanation:

These come directly from my textbook, so I'm not sure if your teacher will accept this kind of work.

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Construct: An angle congruent to angle ABC

Procedure:

1. Draw a ray. Label it ray RY.

2. Using B as center and any radius, draw an arc that intersects ray BA and ray BC. Label the points of intersection D and E, respectively.

3. Using R as center and the same radius as in Step 2, draw an arc intersecting ray RY. Label the arc XS, with S being the point where the arc intersects ray RY.

4. Using S as center and a radius equal to DE, draw an arc that intersects arc XS at a point Q.

5. Draw ray RQ.

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4 0
3 years ago
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weeeeeb [17]

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3 years ago
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