Answer:
p= 2.5
q= 7
Step-by-step explanation:
The lines should overlap to have infinite solutions, slopes should be same and y-intercepts should be same.
Equations in slope- intercept form:
6x-(2p-3)y-2q-3=0 ⇒ (2p-3)y= 6x -2q-3 ⇒ y= 6/(2p-3)x -(2q+3)/(2p-3)
12x-( 2p-1)y-5q+1=0 ⇒ (2p-1)y= 12x - 5q+1 ⇒ y=12/(2p-1)x - (5q-1)/(2p-1)
Slopes equal:
6/(2p-3)= 12/(2p-1)
6(2p-1)= 12(2p-3)
12p- 6= 24p - 36
12p= 30
p= 30/12
p= 2.5
y-intercepts equal:
(2q+3)/(2p-3)= (5q-1)/(2p-1)
(2q+3)/(2*2.5-3)= (5q-1)/(2*2.5-1)
(2q+3)/2= (5q-1)/4
4(2q+3)= 2(5q-1)
8q+12= 10q- 2
2q= 14
q= 7
Answer:
It's -7
Step-by-step explanation:
The perpendicular bisector theorem gives the statements that ensures
that
and
are perpendicular.
The two statements if true that guarantee
is perpendicular to line
are;
Reasons:
The given diagram is the construction of the line
perpendicular to line
.
Required:
The two statements that guarantee that
is perpendicular to line
.
Solution:
From the point <em>C</em> arcs <em>E</em> and <em>D</em> are drawn to cross line
, therefore;
arcs drawn from the same radius.
is perpendicular to line
, given.
Therefore;
by perpendicular bisector theorem.
Learn more about the perpendicular bisector theorem here:
brainly.com/question/11357763
Answer:
- x = 7
- x = 11
- 5
Step-by-step explanation:
![\frac{1}{7} \times x = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B7%7D%20%20%5Ctimes%20x%20%3D%201)
![7 \times \frac{1}{7} x = 7 \times 1](https://tex.z-dn.net/?f=7%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B7%7D%20x%20%3D%207%20%5Ctimes%201)
![7 \times \frac{1}{7} x = 7](https://tex.z-dn.net/?f=7%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B7%7D%20x%20%3D%207)
![\boxed{\green{x = 7}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cgreen%7Bx%20%3D%207%7D%7D)
◇◇•◇◇
![x \times \frac{1}{11} = 1](https://tex.z-dn.net/?f=x%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B11%7D%20%20%3D%201)
![\frac{1}{11} x = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B11%7D%20x%20%20%3D%201)
![11 \times \frac{1}{11} x = 11 \times 1](https://tex.z-dn.net/?f=11%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B11%7D%20x%20%3D%2011%20%5Ctimes%201)
![11 \times \frac{1}{11} x = 11](https://tex.z-dn.net/?f=11%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B11%7D%20x%20%3D%2011)
![\boxed{\green{x = 11}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cgreen%7Bx%20%3D%2011%7D%7D)
♤♤•♤♤
![1 \div \frac{1}{5} = 1 \times \frac{5}{1}](https://tex.z-dn.net/?f=1%20%5Cdiv%20%20%5Cfrac%7B1%7D%7B5%7D%20%20%3D%201%20%5Ctimes%20%20%5Cfrac%7B5%7D%7B1%7D%20)
![1 \times 5](https://tex.z-dn.net/?f=1%20%5Ctimes%205)
![\boxed{\green{= 5}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Cgreen%7B%3D%205%7D%7D)
69 - 3a = 0
69 - 3a (-69) = 0 (-69)
-3a = -69
-3a(/-3) = -69(/-3)
a = 23
hope this helps