Segment drawn on scale having end points O (0,0) and A 
is a line segment.
O A= ![\sqrt{[\frac{3}{4} -0]^{2} +[\frac{9}{10} -0]^{2}\\\\](https://tex.z-dn.net/?f=%5Csqrt%7B%5B%5Cfrac%7B3%7D%7B4%7D%20-0%5D%5E%7B2%7D%20%2B%5B%5Cfrac%7B9%7D%7B10%7D%20-0%5D%5E%7B2%7D%5C%5C%5C%5C)
O A = 
= 
= 
Now , the same segment actual structure having end points O'(0,0) and B (30,36) is also a line segment.
O'B= 
= 
= 
= 2√549
[Cancelling √549 from numerator and denominator]
So, Actual length = 40 × Length on scale
Answer:
x=-2, y=-3. (-2, -3).
Step-by-step explanation:
-5x+4y=-2
2x-7y=17
-------------------
2(-5x+4y)=2(-2)
5(2x-7y)=5(17)
---------------------
-10x+8y=-4
10x-35y=85
-------------------
-27y=81
y=81/-27
y=-3
2x-7(-3)=17
2x+21=17
2x=17-21
2x=-4
x=-4/2
x=-2
i really cant see what it says sorry
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:32,768
2*4=8*4=32*4=128
So we have to multiply by 4
128*4=512 5th term
512*4=2048 6th term
2048*4=8192 7th term
8192*4=32768 last term
