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katen-ka-za [31]
3 years ago
15

Which equation represents the function graphed on the

Mathematics
1 answer:
Vesna [10]3 years ago
4 0
Since you haven't provided the graph, I'll explain each one and you choose the one suiting your given.
The parent modulus function is:
g(x) = |x|
It is centered at the origin and opens upwards.
A coefficient inside the modulus |x+k| means that the function is shifted along the x-axis
If "k" is positive, the shift will be to the left. If "k" is negative, the shift will be to the right.
A coefficient outside the modulus |x| + h means that the function is shifted along the y-axis
If "h" is positive, the shift will be upwards. If "h" is negative, the shift will be downwards.
Now, let's check each of the options:
g(x) = |x+4| - 2 :
This function is shifted 4 units to the left and 2 units down. It will be centered at (-4,-2). Check the blue graph in the attachment.
g(x) = |x-4| - 2 :
This function is shifted 4 units to the right and 2 units down. It will be centered at (4,-2). Check the black graph in the attachment.
g(x) = |x-2| - 4 :
This function is shifted 2 units to the right and 4 units down. It will be centered at (2,-4). Check the red graph in the attachment.
g(x) = |x-2| + 4 :
This function is shifted 2 units to the right and 4 units up. It will be centered at (2,4). Check the green graph in the attachment.
All 4 graphs are shown in the attached picture.
Hope this helps :)



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Given parallelogram E F G H with diagonals that intersect at point J comma segment E J is congruent to which segment?
guapka [62]

Answer:

Given: A parallelogram EFGH in which diagonals intersect at point J.

To Find: A segment congruent to EJ.

Solution: In parallelogram EFGH

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As we know diagonals of parallelogram bisect each other.

So, EJ=JG and FJ=JH

∵ Solution is EJ≅ JG




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2 years ago
The value of x that satisfies the equation 4/3=x+10/15
kenny6666 [7]

For this case we must find the value of "x" that meets the following equation:

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We subtract \frac {10} {15} on both sides of the equation:

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x =\frac {2} {3}

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