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BabaBlast [244]
3 years ago
8

Which equation represents a line which is parallel to the x-axis?

Mathematics
1 answer:
Allushta [10]3 years ago
3 0

Answer:

y = 8

Step-by-step explanation:

The equation of a line parallel to the x- axis has equation

y = c

where c is the value of the y- coordinates the line passes through.

Then

y = 8 ← is equation of line parallel to the x- axis

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What is the length of segment BC? (4 points)
valkas [14]

Answer:

the answer is 7 units

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Please explain how you came to your answer, need help ASAP. Will give brainliest
shusha [124]
The vertex is the high point of the curve, (2, 1). The vertex form of the equation for a parabola is
.. y = a*(x -h)^2 +k . . . . . . . for vertex = (h, k)

Using the vertex coordinates we read from the graph, the equation is
.. y = a*(x -2)^2 +1

We need to find the value of "a". We can do that by using any (x, y) value that we know (other than the vertex), for example (1, 0).
.. 0 = a*(1 -2)^2 +1
.. 0 = a*1 +1
.. -1 = a

Now we know the equation is
.. y = -(x -2)^2 +1

_____
If we like, we can expand it to
.. y = -(x^2 -4x +4) +1
.. y = -x^2 +4x -3

=========
An alternative approach would be to make use of the zeros. You can read the x-intercepts from the graph as x=1 and x=3. Then you can write the equation as
.. y = a*(x -1)*(x -3)
Once again, you need to find the value of "a" using some other point on the graph. The vertex (x, y) = (2, 1) is one such point. Subsituting those values, we get
.. 1 = a*(2 -1)*(2 -3) = a*1*-1 = -a
.. -1 = a
Then the equation of the graph can be written as
.. y = -(x -1)(x -3)
In expanded form, this is
.. y = -(x^2 -4x +3)
.. y = -x^2 +4x -3 . . . . . . same as above
5 0
4 years ago
6/2=4/p= Can u help me with this math problem
vodomira [7]
I hope this helps you


6/2=4/p


3=4/p


p=4/3

p=1,333...
5 0
4 years ago
F plus 4 divided by h= 6
Sergeeva-Olga [200]
F is 8 h is 2 hope that helpssss
7 0
3 years ago
I need help with this problem
cupoosta [38]
\bf \begin{cases}
(n^4)^p=n^{12}\to &n^{4\cdot p}=n^{12}\to 4p=12
\\
&\qquad \uparrow \\
&\textit{same bases, thus}\\
&\qquad \downarrow 
\\
n^3\cdot n^q=n^6\to &n^{3+q}=n^6\to 3+q=6
\end{cases}
\\\\\\
thus\implies 
\begin{cases}
4p=12\implies p=\frac{12}{4}
\\\\
3+q=6\implies q=6-3
\end{cases}
\\\\

\\\\
then\implies p\cdot q=\boxed{?}
7 0
3 years ago
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