Answer:
The answer in the procedure
Step-by-step explanation:
we have
2x-3y=-1 ----> equation A
3x+3y=26 --> equation B
Solve the system by elimination
Adds equation A and equation B
2x-3y=-1
3x+3y=26
----------------
2x+3x=-1+26
5x=25
x=25/5
x=5
Find the value of y
substitute the value of x in equation A or equation B and solve for y
2(5)-3y=-1
10-3y=-1
3y=10+1
y=11/3
Answer:
70 m/s
Step-by-step explanation:
speed = √( g x depth ); where g = 9.8m/s²
given depth = 500m
equation becomes:
speed = √( g x depth )
= √( 9.8 x 500 )
= √( 9.8 x 500 )
=√4900
=70 m/s
Answer:
15n +8 < -10n-12
Step-by-step explanation:
Dy/dx = y/x^2
1/y dy = 1/x^2 dx
int(1/y dy) = int(1/x^2 dx)
ln y + c1 = -x + c2
ln y = -x + c ; c= c2-c1
y = e^(-x+c) = e^c e^-x
y = Ce^-x ; C=e^c
Given:
Tangent segment MN = 6
External segment NQ = 4
Secant segment NP =x + 4
To find:
The length of line segment PQ.
Solution:
Property of tangent and secant segment:
If a secant and a tangent intersect outside a circle, then the product of the secant segment and external segment is equal to the product of the tangent segment.
Subtract 16 from both sides.
Divide by 4 on both sides.
The length of line segment PQ is 5 units.
