The answer is: -3y+9z
This is because you simply switch the sign for every number in the parentheses.
Hope this helps!
Answer:
<em>not</em> a rectangle
Step-by-step explanation:
There are several ways to determine whether the quadrilateral is a rectangle. Computing slope is one of the more time-consuming. We can already learn that the figure is not a rectangle by seeing if the midpoint of AC is the same as that of BD. (It is not.) A+C = (-5+4, 5+2) = (-1, 7). B+D = (1-2, 8-2) = (-1, 6). (A+C)/2 ≠ (B+D)/2, so the midpoints of the diagonals are different points.
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The slope of AB is ∆y/∆x, where the ∆y is the change in y-coordinates, and ∆x is the change in x-coordinates.
... AB slope = (8-5)/(1-(-5)) = 3/6 = 1/2
The slope of AD is computed in similar fashion.
... AD slope = (-2-5)/(-2-(-5)) = -7/3
The product of these slopes is (1/2)(-7/3) = -7/6 ≠ -1. Since the product is not -1, the segments AB and AD are not perpendicular to each other. Adjacent sides of a rectangle are perpendicular, so this figure is not a rectangle.
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Our preliminary work with the diagonals showed us the figure was not a parallelogram (hence not a rectangle). For our slope calculation, we "magically" chose two sides that were not perpendicular. In fact, this choice was by "trial and error". Side BC <em>is perpendicular</em> to AB, so we needed to choose a different side to find one that wasn't. A graph of the points is informative, but we didn't start with that.
Answer:
Step-by-step explanation:
△ABC is isosceles. So ∠3 = ∠4
given 1 ≅ 3.
therefore, 1 ≅ 4
alternate interior angles are equal if only AB//CD
-2 ( 6 + s ) >_ -15 - 2s
We distribute the left side of the inequality.
-12 - 2s >_ -15 - 2s
Add 2s on both sides.
They cancel out, and we’re left with:
-12 >_ -15
This is surprisingly a true statement, so s = 0 in order for us to end up with this again.
Hopefully it’s right. Good luck!
Answer:
The radius of the circle is 3 unit.
Step-by-step explanation:
Given : The diameter of a circle is 6 units.
To find : What is the radius of the circle?
Solution :
The radius of the circle is ![r=\frac{d}{2}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D)
Where, d is the diameter d=6 units
Substitute the value,
![r=\frac{6}{2}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B6%7D%7B2%7D)
![r=3](https://tex.z-dn.net/?f=r%3D3)
Therefore, the radius of the circle is 3 unit.