1. 3x - 2y = 0 => y = 3/2x
4x + 2y = 14<=> 4x + 2*(3/2x) = 14<=> 7x = 14<=> x = 2 & y = 3
2.3p + q = 7=> q = 7 - 3p
2p - 2q = -6<=> 2p - 2*(7-3p) = -6<=> 2p - 14 + 6p = -6<=> 8p = -6 + 14 = 8<=> p = 1 & q = 4
3.3x - 2y = 1=> x = (1+2y)/3
8x + 3y = 2<=> 8*(1+2y)/3 + 3y = 2<=> 8*(1+2y)/3 - 2 = -3y<=> 3*(8*(1+2y)/3 - 2) = -3*(3y)<=> 8*(1+2y) - 6 = -9y<=> 8 + 16y - 6 = -9y<=> 2 = -25y<=> y = -2/25 & x = 7/25
Answer:
Hypothenus = 22
Step-by-step explanation:
From the question given above, we were told that the triangles are congruent (i.e same size). Thus,
AC = EF
BC = DE
To obtain the length of each Hypothenus, we shall determine the value of y and x. This can be obtained as follow:
For y:
AC = y + 3
EF = 2y + 1
AC = EF
y + 3 = 2y + 1
Collect like terms
3 – 1 = 2y – y
2 = y
y = 2
For x:
BC = 5x + 7
DE = 6x + 2y
y = 2
DE = 6x + 2(2)
DE = 6x + 4
BC = DE
5x + 7 = 6x + 4
Collect like terms
7 – 4 = 6x – 5x
3 = x
x = 3
Finally, we shall determine the length of each Hypothenus. This can be obtained as follow:
Hypothenus = BC
Hypothenus = 5x + 7
x = 3
Hypothenus = 5x + 7
Hypothenus = 5(3) + 7
Hypothenus = 15 + 7
Hypothenus = 22
OR
Hypothenus = DE
DE = 6x + 2y
y = 2
x = 3
Hypothenus = 6(3) + 2(2)
Hypothenus = 18 + 4
Hypothenus = 22
Answer: Distance between line and point =
4√5 -3/2√10
Step-by-step explanation:
Distance between the line is
= √ ((9-0)²+(0+1)²)
= √ (89+1)
= √90
= 3√10
Half of the line = 3/2√10
Distance of one side of the line and the point.
= √((9-1)²+(0-4)²)
= √((8)²+(-4)²)
=√64+16
= √80
= 4√5
Distance between line and point =
4√5 -3/2√10
Answer:
90x^2+134x+48
Step-by-step explanation:
To get the area of a classroom you just multiply the length times the width
Step-by-step explanation:
Given: 
and 
We can solve for f(x) by writing
Let 
Then
We know that f(0) = 0 so we can find the value for k:
Therefore,
