Since XY is the perpendicular bisector of JK, and when we say perpendicular, it is a line which cuts a line segment into two equal parts at 90°. So the statements that must be true about it would be all of the above: XY and JK form four right angles, P is the midpoint of XY, m D. XP=YP, JP=KP and XY perpendicular to JK. Hope this answer helps.
Answer:
3 2/3
Step-by-step explanation:
Answer:
n = 5/22
Step-by-step explanation:
That's a final solution.
Answer:
![y = -\frac{3}{2} x + 4](https://tex.z-dn.net/?f=y%20%3D%20-%5Cfrac%7B3%7D%7B2%7D%20x%20%2B%204)
Step-by-step explanation:
First, let us find the gradient of AB:
Gradient of AB = ![\frac{-1-3}{-1-5}](https://tex.z-dn.net/?f=%5Cfrac%7B-1-3%7D%7B-1-5%7D)
= ![\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D)
We also need to know that The <em>product of gradients which are perpendicular to each other is -1</em>. Using this idea, we can find the gradient of the perpendicular bisector:
(Gradient of perpendicular bisector)(
) = -1
Gradient of perpendicular bisector = ![-\frac{3}{2}](https://tex.z-dn.net/?f=-%5Cfrac%7B3%7D%7B2%7D)
Now, we need to know at which coordinates the perpendicular bisector intersects AB. <em>A perpendicular bisector bisects a line to two equal parts</em>. Hence the <em>coordinates of the intersection point is the midpoint of AB</em>. Thus,
Coordinates of intersection = (
,
)
= ( 2, 1 )
Now, we can construct our equation. The equation of a line can be formed using the formula
where
is the gradient and the line passes through
. Hence by substituting the values, we get:
![(y - 1) = -\frac{3}{2} (x - 2)](https://tex.z-dn.net/?f=%28y%20-%201%29%20%3D%20-%5Cfrac%7B3%7D%7B2%7D%20%28x%20-%202%29)
![y - 1 = -\frac{3}{2} x + 3](https://tex.z-dn.net/?f=y%20-%201%20%3D%20-%5Cfrac%7B3%7D%7B2%7D%20x%20%2B%203)
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Answer:
<h2>v =3</h2>
Step-by-step explanation:
![v=\log _7\left(343\right)\\\\\mathrm{Rewrite\:}343\mathrm{\:in\:power-base\:form:}\quad 343=7^3\\\\\log _7\left(7^3\right)=3\log _7\left(7\right)\\\\v=3\log _7\left(7\right)\\\\\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1\\\\\log _7\left(7\right)=1\\\\v=3\times \:1\\\\=3](https://tex.z-dn.net/?f=v%3D%5Clog%20_7%5Cleft%28343%5Cright%29%5C%5C%5C%5C%5Cmathrm%7BRewrite%5C%3A%7D343%5Cmathrm%7B%5C%3Ain%5C%3Apower-base%5C%3Aform%3A%7D%5Cquad%20343%3D7%5E3%5C%5C%5C%5C%5Clog%20_7%5Cleft%287%5E3%5Cright%29%3D3%5Clog%20_7%5Cleft%287%5Cright%29%5C%5C%5C%5Cv%3D3%5Clog%20_7%5Cleft%287%5Cright%29%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Alog%5C%3Arule%7D%3A%5Cquad%20%5Clog%20_a%5Cleft%28a%5Cright%29%3D1%5C%5C%5C%5C%5Clog%20_7%5Cleft%287%5Cright%29%3D1%5C%5C%5C%5Cv%3D3%5Ctimes%20%5C%3A1%5C%5C%5C%5C%3D3)