N^2=24
n=√24
n=√(4*6)
n=2√6
n≈4.90 (to the nearest thousandth)
<h3>
Answer: -4</h3>
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Explanation:
We can pick any two rows from the table to get the (x,y) points needed to find the slope.
Let's say we pick the second and third rows
Subtract the y values: 14-6 = 8
Subtract the x values in the same order: 1-3 = -2
Divide the differences: 8/(-2) = -4
The slope is -4
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You can use the slope formula
Let's say the points are (x1,y1) = (1,14) and (3,6)
m = (y2-y1)/(x2-x1)
m = (6-14)/(3-1)
m = -8/2
m = -4
It's the same basic idea as the previous section. You subtract the y values together (y2-y1) and the x values together (x2-x1) and divide the differences to get m. The order of subtraction doesn't matter as long as you stay consistent. If you do something like y2-y1 and x1-x2, then you'll get the wrong slope value.
Choices A and B are correct. To find the equivalent expressions, you would do the distributive property for the choices to
get rid of the parentheses. You would then see which choices match up with 15 - 5x. Hope this helps!
A.) -5x + 15 (yes)
B.) 15 - 5x (yes)
C.) -15 + 5x (no)
D.) 5 - x (no)
Answer:
5,000
Step-by-step explanation:
can i have brainliest i knew take the 5 away from 50 and take all 0's away from 50 and 100 to get the Answer
5 frist then 000 to get 5,000 i'm smart
Please Dont Copy This Or i will report you
Answer:
F is the midpoint of AA' because Line E G bisects AA'
Step-by-step explanation:
<u><em>The picture of the question in the attached figure</em></u>
we know that
When reflecting a figure, the image is congruent to the pre-image.
A reflection maps every point of the pre-image to an image across a fixed line. The fixed line is called the line of reflection.
The distance of each point of the pre-image from the line of reflection will be the same as the distance of each point of image from the line.
In this problem the line of reflection is E G
The distance A F is equal to the distance A'F
Point F is the midpoint segment A A'
so
Line segment E G bisects segment A A'
therefore
F is the midpoint of A A' because Line E G bisects A A'
