Answer:
300 SUVs were sold.
210 Passenger cars were sold.
Step-by-step explanation:
Let S denote the amount of SUVs sold and let P denote the amount of passenger cars sold.
The total amount of vehicles sold was 510, so:
90 more SUVs were sold than passenger cars. In other words:
We know have a system of equations. We can solve it buy substituting the second equation into the first. Thus:
Combine like terms:
Subtract 90 from both sides:
Divide both sides by 2:
So, the 210 passenger cars were sold.
This means that 210+90=300 SUVs were sold.
And we're done!
Answer:
4x-3y+12=0 is required eqn
Step-by-step explanation:
Let, 3,8 be x1,y1
9,16 be x2,y2
Now,
Eqn of st.line is y-y1 = y2-y2/x2-x1 * (x-x1)
y-8 = 16-8/9-3 * (x-3)
y-8 = 4/3*(x-3)
3y-24 = 4x-12
4x-3y+12=0 is required eqn
Find the scale between sides xz and gh
18/4 = 4.5
Multiply the length of fh by the scale to find x
6 x 4.5 = 27
x = 27
Answer:
a)13x+3
b) -3x-13
c)4x^2-24
d)-3x^2 +3x
Step-by-step explanation:
a) combine like terms
5x-3+8x+6 so combine the -3 and +6 and 5x and 8x
So 13x+3
b)There is an invisible one in front of (2x-5) and an invisible -1 in front of (5x+8). Distribute these.
2x -5 - 5x -8
So -3x-13
c) Distribute the four to everything in the brackets.
4(x^2-6) becomes 4x^2-24
d) Distribute:
-3x (x-1) becomes -3x^2 +3x
Answer:
a) True
b) False
c) False
d) False
e) True
Step-by-step explanation:
a) Each basis of V has four vectors. Then any set of 5 vectors must be linear dependent (LD).
b) Suppose that 
is a basis of V. Considere the set 
where 
are scalars. The set has 5 vectors but 
because 
is not belong to A and is linear independent of 
c) Suppose that is a basis of V. Considere the set 
where 
are scalars. A has four nonzero vectors but isn't a basis because is a LD set.
d) Suppose that is a basis of V. Considere the set 
where are scalars. A has 3 nonzero vectors but isn't a basis because is a LD set.
e) Since any basis of V must have 4 elements, then a set of three vectors cannot generate V.
