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evablogger [386]
3 years ago
10

Please please help me

Mathematics
1 answer:
galben [10]3 years ago
4 0

Answer:

yes the graph reflects over the y-axis because if you see in the y axis the line show that it is the pont to the axis

hope it helps

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BRAINLIEST ANSWER FOR FIRST Please help with the problem below
Sidana [21]
Answer:

m\:]

Step-by-step explanation:

The given inequality is ;

\frac{2m+3}{2}-\frac{17}{4}\:

Multiply through by 8 to clear the fraction;

4(2m+3)-2(17)\:

Expand;

8m+12-34\:

Group similar terms;

8m-m\:

7m\:

m\:
7 0
3 years ago
Read 2 more answers
I’m so bad at this ughhhhhhhhhhhhhhhhhhh
Alexus [3.1K]

Answer:

a)6

b)v^2-u^2/2a

Step-by-step explanation:

a)Given that,

u=12

a= -6

s=9

We know that,

v^2=u^2+2as

or, v^2=(12)^2+2×(-6)×9

or,v^2=144-108

or,v^2=36

or, v=√36

∴v=6

         (Ans)

b)If ,v^2=u^2+2as

then, u^2+2as=v^2

or,2as=v^2-u^2

or, s=v^2-u^2/2a

∴s=v^2-u^2/2a

Hope ya find it helpful.. Thanks a lot...

6 0
3 years ago
Graph y= x^2+3x+2. identify the vertex and axis of symmetry
dlinn [17]
Axis or sym. is -1.5. vertex is ( -.25,-1.5)

5 0
3 years ago
Solve for j. -1j+1&gt;2<br> Please help.
prohojiy [21]
You can use photomath for this
6 0
3 years ago
How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

3 0
3 years ago
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