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sveta [45]
3 years ago
15

What is the slope-intercept equation of the line below?

Mathematics
1 answer:
laila [671]3 years ago
7 0

Answer:

There is no image.

Step-by-step explanation:

Can you add a image or something because You did not give a "line below"

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How do I get the answer to this math problem n + 7 < -3 what is the formula for the math problem
laiz [17]
If I understand the question, you have to get n alone. in this case, you have to subtract 7 from both sides, resulting in n (is less than) -10
8 0
3 years ago
6 red marbles out of 36 marbles
Andreyy89

Answer:

1/6

Step-by-step explanation:

P(red) = red/total

             =6/36

             =1/6

6 0
3 years ago
607351+57299-57389=? ANSWER REALLY FAST
castortr0y [4]

Answer:

607,261

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Instructions: Find the missing side lengths. Leave your answers as radicals in simplest form.
ivanzaharov [21]

Answer:

x = 6

y = 3

Step-by-step explanation:

Give triangle is a right triangle.

By applying sine ratio in the given triangle,

sin(60)° = \frac{\text{Opposite side}}{\text{Hypotenuse}}

\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{x}

x = 6

Now we take the cosine ratio in the given triangle,

cos(60)° = \frac{\text{Adjacent side}}{\text{Hypotenuse}}

\frac{1}{2}=\frac{y}{x}

y = \frac{x}{2}

y = \frac{6}{2}=3

Therefore, x = 6 and y = 3 will be the answer.

3 0
3 years ago
Derive the equation of the parabola with a focus at (3,1) and a directrix of y = 5
serg [7]
So hmmm  check the picture below

so... the vertex is "p" distance from the focus and the directrix, thus, the vertex is really half-way between both

in this case, 2 units up from the focus or 2 units down from the directrix, and thus it lands at 3,3

now, the "p" distance is 2, however, the directrix is up, the focus point is below it, the parabola opens towards the focus point, thus, the parabola is opening downwards, and the squared variable is the "x"

because the parabola opens downwards, "p" is negative, and thus, -2

now, let's plug all those fellows in then

\bf \begin{array}{llll}
(x-{{ h}})^2=4{{ p}}(y-{{ k}})\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-----------------------------\\\\

\begin{cases}
h=3\\
k=3\\
p=-2
\end{cases}\implies (x-3)^2=4(-2)(y-3)\implies (x-3)^2=-8(y-3)
\\\\\\
-\cfrac{(x-3)^2}{8}=y-3\implies \boxed{-\cfrac{1}{8}(x-3)^2+3=y}

5 0
4 years ago
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